If one of the $x$ or $y$ is $0$ then the other one must be $0$. Take mod $x$ give us $x|y$ Take mod $y$ give us $y|x$ So we easily conclude $x=y$ $\Longrightarrow x^4-2x^3-x^2+2x=(x-2)(x-1)x(x+1)$ , which mean $(x,y)=(0,0)=(1,1)=(2,2)$ where $x,y$ are non-negetive.

If $0\in\{x,y\}$ then $\boxed{(x,y)=(0,0)}$, which works. Checking$\pmod x$, we have $x\mid y$, and$\pmod y$ gives $y\mid x$. Then $x=y\ne0$, so we are left with $x^4+2x=x^2+2x^3$ or $x(x-1)(x+1)(x-2)=0$, hence $\boxed{(x,y)=(1,1),(2,2)}$ also work.

funny problem First, note that if either $x$ or $y$ are $0$, then $(0,0)$ works. Now take $\pmod x$ and $\pmod y$ to get $y \equiv 0 \pmod x$ and $x \equiv 0 \pmod y$, so $x=y$. Thus the equation reduces as $$x^4+2x=x^2+2x^3\rightarrow x(x-1)(x+1)(x-2)=0\rightarrow x=1,2,$$so the only working pairs are $(x,y)=(0,0),(1,1),(2,2)$.