Disclaimer: The solutions we've shared are just one exciting approach, and there are surely many other wonderful methods out there. We’d love to hear your alternative solutions in the community thread below, so let's keep the creativity flowing!

The important observation is that only one element is y-free and only one element is x- free. Since RHS is divisble by y, LHS must be too. 

So we have,

$y|x^3y + x+y $

$\ \Rightarrow y | x$ 

Which means $y<=x$ or $x=0$.

If $x=0$, then $y=0$.

If $ y<=x$, By similar process we get $x<=y$. Combining the two equations we get $y= x$. 

Now,  

$x^4 + 2x = x^2 + 2x^3$

$\Rightarrow x^3 -2x^2 - x + 2 = 0$

$\Rightarrow (x-2)(x+1)(x-1)= 0$

The three solutions we get from here are, $x=1,-1,2$. Due to non negativity, $-1$ is invalid.

So, at last we have $x = 0,1,2$.

So our total solution set is $(x,y)=(0,0),(1,1),(2,2)$