Sum of integer squares

Above equation can be written as

$y^2(1+2x^2) = 48(1+2x^2) + 816$ $(y^2-48)(2x^2+1) = 816 = 2^4×3×17$

Now, as $2x^2 + 1$ is odd, it is either equal to $3$ or $17$ or $51$ .

Only $3$ and $51$ satisfy, so $x = 1$ or $x = 5$ as in $17$ , $x = 2\sqrt2$ is not integer.

So,

$(y^2-48)(51) = 816 \text{ or } (y^2 - 48)3 = 816$

$y^2 - 48 = 16 \text{ or } y^2 - 48 = 273$

$y = 8 \text{ or } y = \sqrt{321}$

As $y$ is integer, therefore $y = 8$ and $x = 5$ are solutions

Therefore,

$x^2 + y^2 = 64 + 25 = 89$

$y^2=16\left (3+\dfrac {51}{2x^2+1}\right )$

Since $y$ is an integer, $2x^2+1\leq 51\implies |x|\leq 5$

So $|x|$ can be $0,1,2,3,4,5$

Of these, only $|x|=5$ yields an integral value of $y$ , for which $y^2=64$

Hence, $x^2+y^2=25+64=\boxed {89}$ .