We discretize the domain $\Omega = [0,1]^2$ with the step size $\Delta x = \frac{1}{n-1}$. Then, for the largest eigenvalue of the discrete Laplacian it holds
$$\lambda_{\max}(-\Delta)\leq 8(n-1)^2.$$
I tried to find a proof for that, but was not successful. Is there an explanation? Thanks!
Edit: the boundary condition for a solution $m$ is given by $\nu \cdot m = 0$ on $\partial \Omega$, where $\nu$ is the normal field of $\Omega$.
If you are using standard finite differences, you matrix will be "penta-diagonal" and Gerschgorin's theorem can give you an estimate of the spectral radius. Using the most populated rows, you can see that any eigenvalue $\lambda$ satisfies $$ \left|\lambda + \frac{4}{h^2}\right|\leq \frac{4}{h^2}. $$
So, you can conclude that $|\lambda| \leq 8(n-1)^2$ (I'm assuming the solution vanishes at the boundary).