Solving the equation w.r.t. $ m$ we obtain $ m = \frac {3\pm\sqrt {9925 - 26n^2}}{5}$, which means $ 9925 - 26n^2$ must be a perfect square.
Let $ k^2 = 9925 - 26n^2$. So we have $ k^2\equiv_{13}6$, and it's easy to see $ 6$ is not a quadratic residue modulo $ 13$, therefore the equation has no solutions on integers.