Not enough information!

$x^{2}$ - $y^{2}$ = $(x + y)(x - y)$ = $7$ As 7 is prime, either $x - y = 1$ or $x + y = 1$ As $x$ and $y$ are positive integers, $x + y = 1$ is not possible $=> x - y = 1 => x = y + 1 => x + y = y + 1 + y = 2y + 1 = 7 => y = 3 => x = 4$ Therefore, $x^{3}$ + $y^{3}$ = $4^{3}$ + $3^{3}$ = 64 + 27 = 91

Well, the only pair of squares which difference is 7 is 4 and 3. Plug it into $x^3+y^3$ and you get $64+27=\boxed{91}$

How to solve it algebraically??

4^2-3^2 = 7, therefore 4^3+3^3=64+27=91