Unknowns as exponents

First note that $z$ must be odd as $2^x$ is even and $3^y$ is odd. Using the fact that odd squares are $1\pmod 4$ we get that $x=1$ and $y$ is odd, or $x\geq 2$ and $y$ is even. If $x$ is odd then $2^x + 3^y \equiv (-1)^x\equiv -1 z^2\pmod 3$ which gives a contradiction as squares are always $0$ or $1$ modulo $3$ . Hence, $x$ is even and as $x\neq 1$ we also have that $y$ is even.

Write $x = 2n$ and $y = 2m$ so that $(2^n)^2 + (2^m)^2 = z^2$ . Integers $a$ , $b$ and $c$ with the property that $a^2 + b^2 = c^2$ are called Pythagorean triples and we know that every triple $(a,b,c)$ can be written in the form $(s^2-t^2, 2st, s^2+t^2)$ with $s,t\in\mathbb N$ . So we can write $2^n = 2st$ , $3^m = s^2-t^2$ and $z = s^2+t^2$ (note that $2^n$ is even and $3^m$ is odd so we must have $(a,b) = (3^m,2^n)$ ).

From $2^n = 2st$ we get $s = 2^k$ and $t = 2^\ell$ for $k,\ell\in\mathbb N$ . As $3^m = 2^{2k} - 2^{2\ell}$ is even, either $k = 0$ or $\ell = 0$ , and because $k > \ell$ (otherwise $s^2-t^2<0$ ) we have $\ell = 0$ so $t = 1$ . This means that $s = 2^{n-1}$ . Now we have $3^m = (2^{n-1})^2 - 1 = (2^{n-1} - 1)(2^{n-1} + 1)$ . Both factors have to be powers of $3$ , but they are not equal modulo $3$ . So the smallest factor has to be $1$ so $2^{n-1} - 1 = 1$ so $n = 2$ . This gives $s = 2$ and so $m = 1$ and $z = 5$ . Hence, the only solution is $x = 4$ , $y = 2$ and $z = 5$ so the sum is $4+2+5 = 11$ .