Hypotenuse = $512 \sqrt 2$

We can use a method similar to infinite descent, though this one only requires a finite number of steps, to show that the only integer solution to the given equation is $(512,512)$ .

Note that $m$ and $n$ cannot have opposite parity, else $m^2+n^2$ would be odd. Also, if $m$ and $n$ are both odd, then $m^2$ and $n^2$ are both $\equiv 1$ mod $4$ (see note at bottom), so $m^2 + n^2 \equiv 2$ mod $4$ , but $2^{19} \equiv 0$ mod $4$ . So $m$ and $n$ must both be even.

Say $m=2p$ and $n=2q$ for some $p<m, q<n$ and $p,q \in \mathbb{N}$ . Then we have

$\begin{aligned} m^2 + n^2 &= 2^{19} \\ (2p)^2 + (2q)^2 &= 2^{19} \\ 4p^2 + 4q^2 &= 2^{19} \\ p^2 + q^2 &= 2^{17} \end{aligned}$

We now have essentially the same equation we started with, but with the power of $2$ decreased by 2. Everything we said about $m, n$ and $2^{19}$ still holds for $p, q$ and $2^{17}$ , so we can replace $p$ with $2s, q$ with $2t$ , getting the equation $s^2 + t^2 = 2^{15}$ , and so on.

We can continue this process until we get a power of $2$ low enough that it is no longer $\equiv 0$ mod $4$ , which of course happens when we reach $2^1$ . At that point the only possible solution to the resulting equation is $1^2 + 1^2 = 2^1$ . Then the two variables in the previous cycle, call them $x$ and $y$ , would have to be $x = 2 \cdot 1 = 2$ and $y = 2 \cdot 1 =2$ ; the variables in cycle before that one, call them $u$ and $v$ , would have to be $u = 2x = 2^2$ and $v = 2y = 2^2$ ; and so on, going back up through our chain.

It took nine cycles to reach the end of our chain, so it will take nine steps to go back up; our only solution for $(m,n)$ is $(2^9,2^9)$ and our answer is $2^9 + 2^9 = 2^{10} = \boxed{1024}$

Note: to see that $n \equiv 1$ mod $2 \implies n^2 \equiv 1$ mod $4$ , let $n = 2k + 1$ , then $n^2 = 4k^2 + 4k + 1$ , and the result follows.

I'm certain an induction proof would be quite straightforward here as well, but given the choice between an induction proof and a non-induction proof, I always prefer the latter.

using $Mathematica$

PowersRepresentations[2^19,2,2]

returns only one pair: {{512, 512}}