As in my title, I was working on the problem that $m^2-n^2$ and $m^2+n^2$ cannot be both perfect squares. Currently I figured out that m should be odd and n should be even.
The process for this part is that: I assume that $gcd(m,n)=1$. If both of them are odd, we have $m^2+n^2$ to be congruent to 2 modulo 4, which is not possible given that 2 is not a quadratic residue.
However I'm stuck here. I struggled to find any more contradictions. Yet I couldn't.
You forgot to say $n \not= 0$.
Set $m^2 - n^2 = a^2$ and $m^2 + n^2 = b^2$. Then $a^2$, $m^2$, and $b^2$ form an arithmetic progression of $3$ positive integer squares with common difference equal to the perfect square $n^2> 0$. Dividing by $n^2$, the sequence $(a/n)^2$, $(m/n)^2$, and $(b/n)^2$ is an arithmetic progression of three rational squares with common difference $1$.
Historically, a positive integer $N$ has been called a congruent number when there is an arithmetic progression of three rational squares with common difference $N$, since such an arithmetic progression was called a congruum. Your task amounts to showing $n^2$ is not a congruent number, which is equivalent (using division by $n^2$) to showing $1$ is not a congruent number. That $1$ is not a congruent number is a result going back to Fermat, and showing this is not easy. So I am not all surprised you are getting stuck.
Two equivalent descriptions of a positive integer $N$ being a congruent number are that (i) there is a right triangle with rational side lengths and area $N$, or (ii) the equation $y^2 = x^3 - N^2x$ has a rational solution $(x,y)$ where $y \not= 0$: see Theorem 3.1 here and Corollary 3.3 here. This is how congruent numbers are nearly always defined in accounts of them today, with $1$ not being a congruent number being a consequence of the fact (which is not obvious) that the only rational solutions to $y^2 = x^3 - x$ are $(0,0)$, $(1,0)$, and $(-1,0)$: see Theorem 3.10, Corollary 3.19, and Appendix A here.
I found a way to show that $m^4-n^4$ could not be a perfect square aside than trivial solutions (that $m^2=n^2$ or $n=0$), and because $m^4-n^4=(m^2+n^2)(m^2-n^2)$, this would suggest that $m^2+n^2$ and $m^2-n^2$ may not both be perfect squares.
This solution is from Diophantine Equations by L. J. Mordell on page 17-18 (thanks to @Will Jagy for recommending the book). The process goes as:
We suppose now that none of $x,y,z$ is zero.
Let $x > 0, y > 0, z > 0$ be the primitive solution of $m^4-n^4=k^2$ for which $x$ is the least. Clearly $x$ is odd and then $y$ may be even or odd. Now assume that $m≠1$.
If $y$ is odd, then $x^2 = a^2 + b^2$, $y^2 = a^2 - b^2, z = 2ab$ for some $(a, b) = 1, a > b > 0$, and so $x^2y^2 = a^4 - b^4$. This is a proper solution and since $a < x$, we have a contradiction.
If $y$ is even, then $x^2 = a^2 + b^2$, $y^2 = 2ab$ for some$(a, b) = 1, a > 0, b > 0$, and we may suppose that a is even and b is odd.
Then $a = 2p^2$, $b = q^2$, for some odd $q$ and $(p, q) = 1, p > 0, q > 0$. and so $x^2 = 4p^4 + q^4, y = 2pq$.
Then $p^2 = rs, q^2 = r^2 - s^2$ for some $(r, s) = 1, r > s > 0$ due to $x^2=(2p^2)^2+(q^2)^2$.
Then $r = u^2, s = v^2$ for some (u, v) = 1, u > v > 0 due to $rs$ being a perfect square and $(r,s)=1$.
This would cause $u^4-v^4=q^2$ with $u<x$, which contradicts that x is the smallest positive integer to be able to do so.
Therefore no such solution exists. $m^4-n^4$ cannot be a perfect square and we get the desired result.
