Rearranging the given equation leads to a quadratic equation with respect to $n$: $$n^2 + 3n +(5-m^2)=0$$ Therefore $$n = \frac{-3\pm\sqrt{3^2 - 4\times(5-m^2)}}{2}$$ Because $n$ is an integer, therefore $3^2 - 4\times(5-m^2)$ must be a square number. Let $$3^2 - 4\times(5-m^2) = k^2 \implies 4m^2 - k^2 = 11$$ where $k$ is a positive integer.This equation can be solved using the difference of square method: $$(2m+k)(2m-k)=11$$ Hence of the following must hold: $$ \left\{ \begin{array}{rcl} 2m+k &=& 11\\ 2m-k &=& 1\\ \end{array} \right. \implies m=3, k = 5 \implies n = 1, -4 $$ In conclusion, all the solutions are $$(m,n)=\boxed{(3,1), (3,-4)}$$