Either $x+y=0$ in which case $(x, -x)$ is a solution, or we can divide by $x+y$ on both sides to get $x^2-xy+y^2=x+y$ or $(x-1)^2+(y-1)^2+(x-y)^2=2$. The only set of squares adding to $2$ is $(1, 1, 0)$. Thus either $x=y$, or one of $x, y$ is $1$. If $x=y$ then $|x-1|=1$ so $(0, 0)$ and $(2, 2)$ work. If $x=1$ then $|y-1|=1$ so $(1, 2)$ and $(1, 0)$ work. Similarly, $(2, 1)$ and $(0, 1)$ work. So all solutions are $(x, -x), (2, 2), (1, 2), (1, 0), (2, 1), (0, 1)$.