Answer to title question is because they’re not the same thing. Let’s provide a review of three definitions, that of algebra over a field, next a Lie algebra, and finally a Poisson algebra. For the sake of concreteness, you may want to assume the field is always $\Bbb{R}$.
Definition 1. (Algebra over a field)
An algebra $A$ over a field $\Bbb{F}$ is by definition a vector space $A$ over the field $\Bbb{F}$, together with a $\Bbb{F}$-bilinear map $A\times A\to A$, typically denoted $\cdot$, and called the algebra product on $A$ or simply product on $A$.
We say $A$ is associative if for each $a,b,c\in A$, we have $(a\cdot b)\cdot c=a\cdot(b\cdot c)$. We say $A$ is commutative if for each $a,b\in A$, we have $a\cdot b=b\cdot a$.
Finally a notational remark: it is common to write simply $ab$ instead of $a\cdot b$, if the meaning of $\cdot$ is clear from the context.
So, in an algebra $A$, you need a notion of addition, scalar multiplication, and a “product”. Examples include the field $\Bbb{F}$, or more non-trivially, the collection of say all continuous real-valued functions on $[0,1]$, $C([0,1])$ (clearly we can add, scalar multiply, and multiply functions… this gives an associative, commutative algebra over $\Bbb{R}$). More generally, continuous functions on a topological space, or smooth functions on a manifold are examples. Another example is $A=M_{n\times n}(\Bbb{F})$, the space of all matrices, with the usual vector space structure and $\cdot$ being matrix multiplication. I should remark that many authors include associativity as part of the definition, and if they intend something more general, they’ll say something like a “non-associative algebra”.
Definition 2. (Lie algebra)
A Lie algebra (in the abstract sense) over a field $\Bbb{F}$ is by definition a vector space $L$ over $\Bbb{F}$ together with the choice of a bilinear map $[\cdot,\cdot]:L\times L\to L$, called the Lie bracket, such that
- $[\cdot,\cdot]$ is alternating: for all $x\in L$, $[x,x]=0$.
- $[\cdot,\cdot]$ satisfies the Jacobi identity: for all $x,y,z\in L$, $[x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0$.
So, you can say that a Lie algebra is an algebra over $\Bbb{F}$, whereby the “product” $[\cdot,\cdot]$ is alternating, and satisfies the Jacobi identity. Note that $[\cdot,\cdot]$ is not associative in general, which is why one typically abstains from calling it a “product”, and simply refers to it as a “bracket operation”. Also, note that the alternating condition implies skew-symmetry: $[x,y]=-[y,x]$, because for any $x,y\in L$, we have
\begin{align}
0&=[x+y,x+y]=[x,x]+[x,y]+[y,x]+[y,y]=0+[x,y]+[y,x]+0=[x,y]+[y,x].
\end{align}
The converse (i.e skew-symmetry implies alternating) is true for a field of characteristic not equal to $2$ (e.g $\Bbb{R}$ or $\Bbb{C}$, which are the cases of interest in physics).
Perhaps the most common example of a Lie algebra is to start with an associative algebra $(A,\cdot)$, and define the bracket $[\cdot,\cdot]:A\times A\to A$ by setting $[x,y]:=xy-yx$. This is easily checked to be bilinear, alternating, and satisfies the Jacobi identity. We call this the (commutator) Lie bracket on $A$ induced by the product on $A$.
Given this example, it is natural to wonder if the converse holds, namely given a Lie algebra $(L,[\cdot,\cdot])$ over a field $\Bbb{F}$, does there exist a bilinear map $\cdot:L\times L\to L$ which makes $(L,\cdot)$ into an associative algebra such that the given Lie-bracket $[\cdot,\cdot]$ equals the commutator Lie bracket induced by the product $\cdot$. Or more succinctly, are Lie brackets and commutators the same thing. The answer unfortunately is no; see Does every Lie algebra come from commutator of some associative product operation? This is why you need to deal with more abstract stuff than just commutators; in more poetic terms, you’re “cutting out the middle man”. I should remark that while the details of that particular answer may be a little too involved, atleast heuristically you should expect the answer to be no, because in general, on a vector space there is no natural associative product $\cdot$ lying around. So if all you know is $(L,[\cdot,\cdot])$, then where would you even come up with a $\cdot$ from in order to write down the equation $[x,y]=x\cdot y-y\cdot x$?
Lastly, let’s talk about Poisson brackets
Definition 3. (Poisson Algebra)
A Poisson algebra over a field $\Bbb{F}$ is a triple $(P,\cdot,\{\cdot,\cdot\})$ such that $(P,\cdot)$ is an associative algebra over $\Bbb{F}$, $(P,\{\cdot,\cdot\})$ is a Lie-algebra over $\Bbb{F}$ and furthermore, $\{\cdot,\cdot\}$ acts as a derivation: for all $x,y,z\in P$, we have $\{x,y\cdot z\}=\{x,y\}\cdot z+y\cdot\{x,z\}$. We refer to $\{\cdot,\cdot\}$ as the Poisson bracket on $P$.
From the axioms, you can show that $\{\cdot,\cdot\}$ acts as a derivation in the other slot as well: $\{x\cdot y,z\}=x\cdot\{y,z\}+\{x,z\}\cdot y$. The first example one encounters is course starting with some phase space, i.e a symplectic manifold $(M,\omega)$ (which itself the most common examples are cotangent bundles of some configuration manifold $Q$) and looking at all smooth functions $C^{\infty}(M)$ (which is an associative algebra over $\Bbb{R}$ with the usual pointwise product) together with the usual Poisson bracket of functions $\{f,g\}$. On $\Bbb{R}^{2n}$, this gives the familiar formula $\{f,g\}=\frac{\partial f}{\partial q^i}\frac{\partial g}{\partial p_i}-\frac{\partial g}{\partial q^i}\frac{\partial f}{\partial p_i} $ (or maybe there’s a minus sign).
Another very simple example is to start with an associative algebra $(A,\cdot)$, and define the commutator bracket $[x,y]=xy-yx$ as above. We already know this forms a Lie bracket, but even better, this is forms a Poisson bracket too, because
\begin{align}
[x,y]z+y[x,z]=(xy-yx)z+y(xz-zx)=xyz-yzx=[x,yz].
\end{align}
But, one should keep in mind that there may be other interesting Poisson brackets than simply the commutator bracket.
In light of the definitions, it should now be clear that your statement
Lie brackets also share many properties of Poisson's bracket. It is tempting to believe that Lie brackets can also reduce to Poisson brackets
is wrong. It’s the other way around, a Poisson bracket is in particular a Lie bracket, so it is the Poisson bracket which reduces to a Lie bracket (i.e you forget about the associative algebra structure $(P,\cdot)$, and simply focus on $(P,\{\cdot,\cdot\})$).
If you want to read up more, at the group and manifold level, see also Poisson manifolds and Poisson-Lie groups.
Hopefully it’s now clear which is more special than which.
