I don't know whether the OP will be satisfied this answers the original question, but I'd like to offer some context to all this.
A free particle has uniform potential, so without loss of generality $V=0$. The Schrödinger equation then simplifies to $$\dot{\psi}=\frac{i\hbar}{2m}\nabla^2\psi\quad\left(1\right)$$ so $\dot{\rho}=\frac{i\hbar}{2m}\left\{\psi^\ast\nabla^2\psi-\psi\nabla^2\psi^\ast\right\}$. Since probability is conserved, it admits a continuity equation; the probability 3-current $\mathbf{j}$ obeys $\nabla\cdot\mathbf{j}=-\dot{\rho}$. We can choose $\mathbf{j}:=\frac{i\hbar}{2m}\left\{\psi\boldsymbol{\nabla}\psi^\ast-\psi^\ast\boldsymbol{\nabla}\psi\right\}$ (we may add an arbitrary curl to $\mathbf{j}$). If a relativistic 1-particle theory is to work, at the very least a free particle in Minkowski space should be straightforward. In special relativity continuity equations may be written as $\partial_\mu j^\mu=0$. You can check this equation is satisfied by solutions of the mass-$m$ Klein-Gordon equation $$c^{-2}\partial_t^2\psi-\nabla^2\psi+\left(\frac{mc}{\hbar}\right)^2=0,\quad\left(2\right)$$ provided we define $j^\mu\left(\psi\right):=\frac{i\hbar}{2m}\left\{\psi\partial^\mu\psi^\ast-\psi^\ast\partial^\mu\psi\right\}$, which is a natural relativistic upgrade of $\mathbf{j}$. It is therefore natural to suppose a suitable integral of $j^0$ spits out probabilities.
But here we reach a problem. If $\phi,\,\psi$ are equal-mass solutions of the Klein-Gordon equation, we also have a conserved integral called their Klein-Gordon inner product, $$\left\langle\phi,\,\psi\right\rangle_{\text{KG}}:=i\int_{\mathbb{R}^3}\left(\phi^\ast\partial^0\psi-\left(\partial^0\phi^\ast\right)\psi\right)d^3\mathbf{x}.$$ The name is misleading, because this isn't a true inner product; $\left\langle\psi,\,\psi\right\rangle_{\text{KG}}=\frac{2m}{\hbar}\int_{\mathbb{R}^3}j^0\left(\psi\right)d^3\mathbf{x}$ can be negative. Indeed, solutions of Eq. (2) are closed under the operation $\psi\mapsto\psi^\ast$, which multiplies $\left\langle\psi,\,\psi\right\rangle_{\text{KG}}$ by $-1$. Eq. (1) clearly doesn't have an analogous problem (or its usual probability interpretation wouldn't exist). The reason why is that, if we want $\psi\mapsto\psi^\ast$ to send a Schrödinger solution to a Schrödinger solution, we also have to impose $t\mapsto -t$, which also multiplies Klein-Gordon inner products by $-1$.
And the reason why Schrödinger solutions require time reversal and Klein-Gordon solutions don't is because of the parities of the $\partial_t$ exponents. In Schrödinger, the exponent is odd (it's $1$); in Klein-Gordon, the exponent is even (it's $2$). For a classical touchstone, these parities are also why $E=\frac{p^2}{2m}+V$ yields a unique energy but $$E^2=m^2c^4+p^2c^2\,\quad\left(3\right)$$ doesn't.
Nowadays, we know that the way to handle "wrong-sign" solutions of the Klein-Gordon equation is to (i) write solutions as sums of "positive-frequency" and "negative-frequency" parts which interchange under complex conjugation (so the spaces thereof have bases which conjugate to each other's bases) and (ii) say that our space integrals compute differences between numbers of particles and antiparticles.
Let's think now about the Dirac equation. Dirac hoped he could bin negative-energy solutions of Eq. (3) with an equation which, like Schrödinger, was first-order in time. That's how we ended up with $\gamma^\mu\partial_\mu\psi=-im\psi$. This time, to close solutions under $\psi\mapsto\psi^\ast$ we have to append the transformation $x^\mu\mapsto -x^\mu$, which is more than enough to explain the this-time-it-works finding. This time we have not only time reversal but also the spatial equivalent, the parity reversal $\mathbf{x}\mapsto -\mathbf{x}$.
Let's briefly discuss what happens to plane-wave solutions of all three equations. When I conjugate an $\exp i\left(\mathbf{k}\cdot\mathbf{x}-\omega t\right)$ solution (which suffices for the KGE) $\mathbf{k},\,\omega$ change sign. When I then reverse time $\omega$ changes back to its old sign (the Schrödinger requirement), so the only overall change is the sign of $\mathbf{k}$. If I apply a parity transformation as well at the end (Dirac needs this), even this sign change in $\mathbf{k}$ is lost. So actually, our "symmetry" for Dirac solutions does nothing at all!
The last question is what all this has to do with spin and bosons and fermions. The anticommutators $\left\{\gamma^\mu,\,\gamma^\nu\right\}=2\eta^{\mu\nu}$ in $4$-dimensional spacetime require the gamma matrices to be at least $4\times 4$, so the Dirac spinor $\psi$ has at least $4$ components. Dirac realised that the symmetries of the Dirac equation's solutions (not the above "symmetry", some proper ones!) relate these components with a combination of a $2S+1$ spin degeneracy and a matter-antimatter factor of $2$, so $4S+2=4$ and $S=\frac{1}{2}$. Dirac's theory was vindicated not only by predicting the positron, but also by finally explaining spin as a consequence of relativity, whereas before that it was just an empirical fact you had to add to the axioms of quantum mechanics for no apparent reason. This set the stage for later findings concerning spin, such as the spin-statistics theorem. For now, we'll note that a spin-$\tfrac{1}{2}$ Dirac spinor has to be a fermion.
