Problem 848

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

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

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