This is not yet an answer but some progress: I am able to calculate the two positive eigenvalues with great accuracy, as follows:
I perform a similarity transformation on the matrix $M$, given by $M'=S^{-1}MS$ with $$S=\frac{1}{2m}\left( \begin{array}{cccc} 0 & 2 m & -\sqrt{m^2+4 n^2}+m-2 n & \sqrt{m^2+4 n^2}+m-2 n \\ 0 & 0 & -\sqrt{m^2+4 n^2}+m-2 n & \sqrt{m^2+4 n^2}+m-2 n \\ 2 m & 0 & 2 m & 2 m \\ 0 & 0 & 2 m & 2 m \\ \end{array} \right).$$ The matrix $M'$ has the same eigenvalues as $M$ and is given by $M'=M_0+M_1$, with $$M_0=\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{2} \left(-\sqrt{m^2+4 n^2}+m+2 n\right) & 0 \\ 0 & 0 & 0 & \frac{1}{2} \left(\sqrt{m^2+4 n^2}+m+2 n\right) \\ \end{array} \right),$$ $$M_1=\frac{1}{\sqrt{m^2+4 n^2}}\left( \begin{array}{cccc} -2 \sqrt{m^2+4 n^2} & \sqrt{m^2+4 n^2} & -\frac{\sqrt{m^2+4 n^2} \left(\sqrt{m^2+4 n^2}-m+2 n\right)}{2 m} & \frac{\sqrt{m^2+4 n^2} \left(\sqrt{m^2+4 n^2}+m-2 n\right)}{2 m} \\ \sqrt{m^2+4 n^2} & -\sqrt{m^2+4 n^2} & \sqrt{m^2+4 n^2} & \sqrt{m^2+4 n^2} \\ \sqrt{m^2+4 n^2}-2 n & \frac{1}{2} \left(\sqrt{m^2+4 n^2}-m-2 n\right) & -\frac{3}{2} \sqrt{m^2+4 n^2}-\frac{m}{2}+n & -\frac{1}{2} \sqrt{m^2+4 n^2}-\frac{m}{2}+n \\ \sqrt{m^2+4 n^2}+2 n & \frac{1}{2} \left(\sqrt{m^2+4 n^2}+m+2 n\right) & \frac{1}{2} \left(-\sqrt{m^2+4 n^2}+m-2 n\right) & \frac{1}{2} \left(-3 \sqrt{m^2+4 n^2}+m-2 n\right) \\ \end{array} \right).$$ Now we note that for large $m,n$ of order $1/\epsilon$, the matrix $M_0$ is of order $1/\epsilon$ while the matrix $M_1$ is of order unity. We can therefore perform perturbation theory in the small parameter $\epsilon$. To first order the eigenvalues of $M$ are the diagonal elements of $M_0+M_1$, which gives the two negative eigenvalues $\alpha_1=-2$, $\alpha_2=-1$, and the two positive eigenvalues $\beta_+$ and $\beta_-$ given by $$\beta_\pm=\frac{1}{2} \left(\sqrt{m^2+4 n^2}+m+2 n\right)\pm\frac{m-2 n-3 \sqrt{m^2+4 n^2}}{2 \sqrt{m^2+4 n^2}}.$$ This expression for the positive eigenvalues is very accurate, see the plot for $m=n$, where the exact expressions (red and green lines) are almost indistinguishable from the perturbative result (blue and orange).
So to first order in $\epsilon$ one has $\alpha_1+\alpha_2=-3$. The challenge is to prove that higher order corrections are negative.