Yes, eigenvalues of $J^2$ are always of the form $\hbar^2 \ell(\ell+1)$ with $\ell\in \frac{1}{2}\mathbb{Z}_+$ and this can be shown just using the angular momentum algebra. We shall first establish that they are of the form $\hbar^2 \ell(\ell+1)$ for some $\ell \in [0,+\infty)$ and then show that $\ell\in \frac{1}{2}\mathbb{Z}_+$.
First point, $\hbar$ has units of angular momentum. In that case $\hbar^2$ has units of angular momentm squared. In that case if $\mathbf{J}=(J_1,J_2,J_3)$ is an arbitrary angular momentum operator and if $\lambda$ is any eigenvalue of $J^2$ then we must always be able to write $\lambda = \hbar^2 \widetilde{\lambda}$. There is absolutely no loss of generality in doing so, we are just factoring out the dimension.
Second, $J^2$ must have non-negative eigenvalues because it is a sum of three squared Hermitian operators. Since $\hbar^2\geq 0$ then necessarily $\widetilde{\lambda}\geq 0$. But then, any non-negative number can be written as $\widetilde{\lambda}=\ell(\ell+1)$ for some $\ell$.
So we see any possible eigenvalue of $J^2$ for any possible angular momentum will take the form $\hbar^2 \ell(\ell+1)$. The fact that $\ell$ must be integer or half-integer can be shown by studying the angular momentum algebra. Let me outline the proof (I encourage you to fill in the details of the computations):
Define the ladder operators $J_\pm = J_1\pm i J_2$. It then follows from the angular momentum algebra $[J_i,J_j]=i\hbar \epsilon_{ijk}J_k$ that \begin{eqnarray}
[J_3,J_\pm] &=& \pm \hbar J_\pm\tag{1}.
\end{eqnarray}
Now let $|\ell,m,k\rangle$ be the common eigenvectors of $J_3$ and $J^2$, with $k$ an additional degeneracy label. These satisfy $$J^2|\ell,m,k\rangle=\hbar^2 \ell(\ell+1)|\ell,m,k\rangle,\quad J_3|\ell,m,k\rangle = m\hbar |\ell,m,k\rangle\tag{2}.$$ It follows immediately from (1) that $J_\pm$ raises/lowers $m$ by one unit, in other words: $$J_3 J_\pm |j,m,k\rangle = (m\pm 1)\hbar J_\pm |j,m,k\rangle\tag{3}$$
The above observation means that $J_\pm |\ell,m,k\rangle = \alpha |\ell,m\pm 1,k\rangle$. Demanding the state to be normalized we may determine $\alpha$ using the commutation relations. Carrying out this computation $$J_\pm |\ell,m,k\rangle =\sqrt{\ell(\ell+1)-m(m\pm 1)}\hbar |\ell,m\pm 1,k\rangle.\tag{4}$$
Observe that $J^2=J_1^2+J_2^2+J_3^2$. In that case $J^2-J_3^2= J_1^2+J_2^2$ and again, since $J^2-J_3^2$ is a sum of squared Hermitian operators it must have non-negative eigenvalues. This means that the states $|\ell,m,k\rangle$ must satisfy $$m^2\leq \ell(\ell+1)\tag{5}$$
The above observation implies that there must be one $m_{\rm min}$ and one $m_{\rm max}$ such that $m_{\rm min}\leq m\leq m_{\rm max}$ for the allowed states. Since $J_-$ lowers $m$ and $J_+$ raises $m$ it follows that we must necessarily have $J_-|\ell,m_{\rm min},k\rangle=0$ and $J_+|\ell,m_{\rm max},k\rangle=0$, since otherwise the application of these operators would create states which are not allowed in the representation.
This means that $m_{\rm min}$ and $m_{\rm max}$ can be found by these conditions. Inspecting (4) we immediately observe that $m_{\rm min}=-\ell$ and $m_{\rm max}=\ell$. In that case we may construct all the states by applying powers of $J_+$ to $|\ell,-\ell,k\rangle$. What we find is that the allowed values of $m$ are $$m=-\ell,-\ell+1,\dots, \ell-1,\ell\tag{6}$$ and that $\ell$ itself must be integer or half-integer. Indeed to see this last point, observe that there must be an integer $N$ such that $(J_+)^N |\ell,-\ell,k\rangle \propto |\ell,\ell,k\rangle$. Since we know that $(J_+)^N |\ell,-\ell,k\rangle \propto |\ell,-\ell+N,k\rangle$ it is necessary that there be such an integer $N$ with $\ell=N-\ell$ which implies that $2\ell$ must be an integer.
