Find the postive integers such $xy+x+y\mid x^2+y^2-2$ Let $x,y$ be positive integers, and such
$$xy+x+y\mid x^2+y^2-2$$
Find the $x,y$
$$x^2+y^2-2=A(xy+x+y)\Longrightarrow x^2-(Ay+A)x+y^2-Ay-2=0$$

Let $a=x+1$, $b=y+1$. We want $(ab-1)|(a-1)^2+(b-1)^2+2$, which is equivalent to: $(ab-1)$ dividing the numerator of (a-$1)^2+(\frac{1}{a}-1)^2+2$, which is $f(a)=a^4-2a^3-2a+1$. Now let's consider the case where $5 \leq a <b$, the others being solvable by hand. The key observation is that the other factor of $f(a)$ must be of the form $(ac-1)$ for $c<a$. [Otherwise $[a(a+1)-1][a^2+1]>f(a)]$. This means the pair $(a,c)$ is also a solution.
Thus if $(a,b)$ with $5<a<b$ is a solution, then there exists another solution $(c,a)$ with $c<a$. This chain of solutions must stop with the smaller value being at most 4, but for $a \leq 4$, we can check that the corresponding $b$s do not yield other solutions.
