yes (trivial): $x=y=k\in N$

Obviously when $x=y$ Isn't it ? Please edit it to distinct positive integers

I assume that $x$ and $y$ are distinct. Putting $y=kx$ where $k$ is positive integer we have that $ x^2+kx \mid x+k^2x^2 $ from which we deduce $ x+k \mid k^2x+1=x(k^2-x^2)+x^3+1 $ so $ x+k \mid x^3+1 $. So we can now put $k=x^3-x+1$ so pair $(x,y)=(a,a(a^3-a+1))$ works for every integer $a \geq 2$ so there are infinitely many solutions. Of course when $a=1$ also works but then $x$ and $y$ are not distinct

the problem should add "x is not equal to y"

Equation: $\frac{y^2+x}{x^2+y}=q$ Using the solutions of the Pell equation: $p^2-qs^2=\pm1$ Solutions have the form: $y=\pm{p(qp-s)}$ $x=\pm{s(qp-s)}$