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 $x,y$
$$x^2+y^2-2=A(xy+x+y)\Longrightarrow x^2-(Ay+A)x+y^2-Ay-2=0$$
 A: Well, this is the standard contest approach, just an outline. For $A=0,$ the set is a circle and there are integer $(x,y)$ solutions, positive being $(1,1).$ For $A=1,$ the set is an ellipse and there are integer $(x,y)$ solutions, non-negative being $(2,3),$ $(3,2),$ $(2,0),$ $(0,2).$ For $A=2$ there are no integer solutions $\pmod 4,$ where this curve is a parabola.
For $A \geq 3,$ we refer to what is usually called Vieta Jumping on this site. First, two diagrams, for $A=3$ and then $A = 6.$


The approach of Hurwitz (1907) says that, if there are any positive solutions for a certain $A,$ then we may keep reducing $x+y$ by either
$$ x \mapsto A(1+y) - x $$ or
$$ y \mapsto A(1+x) - y, $$ until such time as either: one of (exactly one of, not both) $x,y \leq 0$ or we reach a (positive) fundamental solution (Grundlösung) which satisfies both
$$ 2 y \leq A (1 + x), $$
$$ 2 x \leq A (1 + y). $$
As indicated by the graphs, for $A \geq 3$ the Hurwitz arc lies just outside the first quadrant. However, the branch of the hyperbola we are examining always has $x > -1$ and $y > -1,$  which is to say that there are no integer solutions with $xy < 0.$ Or, with $xy < 0,$ $(x+1)(y+1) \leq 0,$ so $xy+x+y \leq -1,$ while $x^2 + y^2 - 2 \geq 0.$ Oh, if $y=0,$ then $x = (1/2 )(A \pm \sqrt {A^2 + 8});$ this is rational only with $A=1.$
To be really official, we need to show that the Hurwitz arc lies outside the first quadrant when $A \geq 3.$ However, all that is needed is to show that when $x=0$ or $2x = A (1 + y),$ then $y < 0.$ Same by symmetry when $y=0$ or $2y = A (1 + x),$ then $x < 0.$
That is about it, really; Vieta jumping, we cannot have $A \geq 3.$ Much more detail is possible, and I recommend the Hurwitz article to anyone; I have a pdf.
A: 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.
