Eight primes

If $x$ is an odd integer, than $x\equiv1 \mod8$ . Since $h\neq2$ , $a^2+b^2+c^2+d^2+e^2+f^2+g^2\equiv1 \mod8$ . If in the left side there are $k$ number of $2$ 's, then $4k+(7-k)\equiv1 \mod8$ .From these we get $k=6$ (because we get $k\equiv6 \mod8$ ).

Suppose $g\neq2$ . Then $24+g^2=h^2$ and $24=(g+h)(h-g)$ . Since $g$ and $h$ are odd numbers, $g$ and $h$ are $1$ and $5$ or $5$ and $7$ . By calculating we get there is only one solution (supposing $g$ is the one, which is not two):

$2^2+2^2+2^2+2^2+2^2+2^2+5^2=7^2$

So $x=y=24$ , and $xy=24^2=\boxed{576}$