I need to show that there aren't any $m$ and $n$ such that $m^2+n^2$ and $m^2-n^2$ are both squares.
First of all, assume without loss of generality that $m$ and $n$ are co-prime, since otherwise we can divide through by the gcd and get a new pair. Since mod 4 all squares are 0 or 1, the only common solution to both $m^2+n^2$ and $m^2-n^2$ being squares is $m^2\equiv 1$ and $n^2\equiv 0$, so $m$ is odd and $n$ is even.
It also follows easily that $m+n$ and $m-n$ are co-prime: (EDIT: if $p$ divides both of them, then it divides their sum and difference, $2m$ and $2n$, but $p$ is not 2 since $m+n$ is odd, so $p$ divides $m$ and $n$, a contradiction.) Since $m^2-n^2=(m+n)(m-n)$ is a square, both $m+n$ and $m-n$ must be co-prime odd squares.
It seems like I have managed to draw a lot of very strong conclusions from this, but I still can't find any contradictions. Any ideas?