7. The original equation can be transformed into $\left(x^{2}+y^{2}\right)(x+y-8)=8(x y+1)$, then $x, y$ must have the same parity, and $x+y-8$ is even. From this, the integer solutions of the original equation can be found to be $(2,8)$ or $(8,2)$.
7. Find the integer solutions of the equation $x^{3}+x^{2} y+x y^{2}+y^{3}=8\left(x^{2}+x y+y^{2}+1\right)$.