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 $(x,y)=(0,0),(1,1),(2,2)$