Suppose that $\dfrac{m^2+n^2}{mn}=k$. Then we get \[m^2+kmn-n^2=0\]so thus the discriminant is a perfect square. This is \[D=(k^2-4)n^2\]As $n^2$ is a perfect square, we have $k^2\geq 4$, so thus $|k|\geq 2$. Also note if \[k^2-4=m^2\]for $m\geq 0$ then \[k^2-m^2=(|k|+m)(|k|-m)=4\]Then we have \[|k|+m\equiv |k|-m\pmod{2}\]so thus $|k|+m,|k|-m=2$, so thus $k=\pm 2$ are the only solutions.