Factorizing the left side gives $$(x+y)(x^2 -xy + y^2)=(x+y)^2$$

Because both $x$ and $y$ are non-negative, therefore one solution is $$x+y=0\implies (x,y)=(0,0)$$

Cancelling $(x+y)$ on both sides and re-wring it as a quadratic equation of $x$ give: $$x^2 -(y+1)x +(y^2-y)=0$$

In order for this equation has a non-negative solution, it is necessary for its determinant to be a square number. Therefore, $$(y+1)^2-4(y^2 -y)=k^2$$

for some non-negative integer $k$. It follows that $$k^2 +3(y-1)^2 = 4 \implies y=0, 1, 2$$

Accordingly, all the solutions to the original equation are: $$(x, y)=(0, 0), (0, 1), (1, 0), (1, 2), (2, 1), (2,2)$$