Saudi Arabia Mathematical Competitions

Denote by $E(x,y)$ the expression in the problem. We have $$E(x,y)=\frac{(x+y{)}^2}{x(x^2+y^2)-y(x^2+y^2)}=\frac{(x+y{)}^2}{(x-y)(x^2+y^2)}\text{\hspace{1em}(1)}$$ and we can assume that $x>y$.

If $x-y=1$, then $E(x,y)\in \mathbb{Z}$ if and only if $x^2+y^2\mid 2xy$. This is not possible because $x^2+y^2>2xy$.

If $x-y>1$, then $x-y\ge 2$. We have $$0<E(x,y)\le \frac{(x+y{)}^2}{2(x^2+y^2)}<1$$ hence $E(x,y)\notin \mathbb{Z}$.