Some perfect squares

Let $2^x + 3^y = k^2$ for some positive integer $k$ . Now if $x$ were odd, then $2^x$ leaves a remainder of $2$ when divided by $3$ , but $k^2$ can never leave a remainder of $2$ when divided by $3$ .

Therefore $x$ is even , i.e., $x = 2m$ for some positive integer $m$ . Thus

$3^y = k^2 - 2^{2m} = (k-2^m)(k+2^m)$ , so $k-2^m = 3^a$ and $k+2^m = 3^b$ for some non-negative integers $a<b$ with $a+b = y$ . However, $3^b - 3^a = 2^{m+1}$ , and $3$ does not divide $2^{m+1}$ . Hence $a$ must be 0 , and so $b = y$ . So

$k-2^{m} = 1$ , or $k = 2^m + 1$ . Also, $k+ 2^m = 3^y$ , so $2^{m+1} + 1 = 3^{y}$ . Now $2^{m+1}$ must leave a remainder $2$ when divided by $3$ , so $m$ must be even ; say $m = 2p$ for some positive integer $p$ . Thus

$2^{2p+1} + 1 = 3^y$ , and taking modulo $4$ , we get $2 + 1 \equiv (-1)^y {\pmod 4}$ , so $y$ must be even, say $y = 2q$ , for some positive integer $q$ . So $2^{2p+1} = 3^{2q} - 1 = (3^q-1)(3^q+1)$ , and so

$3^q - 1 = 2^r$ and $3^q + 1 = 2^s$ , where $r$ and $s$ are non-negative and $r<s$ . So now

$2^s - 2^r = 2$ , or $2^r(2^{s-r} - 1) = 2$ . Comparing the power of $2$ on both sides, we get $r= 1$ , and so $q = 1$ and thus $p = 1$ . Therefore $\boxed{x = 4}$ and $\boxed{y=2}$ , and their sum is $\boxed{6}$ .