Let $m-n=a$ and $m+n=b$. Then, $b^2-a^2=4mn$. So, we have $$a^2=\frac{b^2-a^2}{b-1}$$$$\implies a^2(b-1)=b^2-a^2$$$$\implies a^2\cdot b = b^2$$$$\implies a^2=b$$$$\implies (m-n)^2=m+n$$Let $m+n=k^2$ for some non-negative $k$. Then, $m-n=\pm k\implies m = n \pm k$. So, $m+n=2n\pm k = k^2$. We split into cases. Case: $2n+k=k^2$ This implies that $n =\frac{k^2-k}{2}$ and $m=\frac{k^2+k}{2}$. Note that $k^2-k$ and $k^2+k$ are always even so $m$ and $n$ are both integers for integer $k$. Plugging in these solutions for our original equation, we find these indeed work for $k\neq 1$. Case: $2n-k=k^2$ This implies that $n=\frac{k^2+k}{2}$ and $m=\frac{k^2-k}{2}$. Note that $k^2-k$ and $k^2+k$ are always even so $m$ and $n$ are both integers for integer $k$. Plugging in these solutions for our original equation, we find these indeed work for $k\neq 1$. $\boxed{(m,n)=(\frac{k^2+k}{2}, \frac{k^2-k}{2}), (\frac{k^2-k}{2},\frac{k^2+k}{2}) \text{for non-negative integer k}\neq 1}$