Find all integer solutions to $ x^3 - y^3 = 3(x^2 - y^2) $ The objective is to find all solutions to
$$ x^3 - y^3 = 3(x^2 - y^2) $$
where $x,y \in \mathbb{Z}$.

So far I've got one pair of solution. Try  $(x, y)=(0,0)$:
$$ 0^3-0^3=3(0^2-0^2) \\
0=0 \qquad \text{equation satisfied}$$
Another try $ (x, y) = (x, x)$ then 
$$ x^3 - x^3 = 3(x^2 -x^2) \\
0 = 0 \qquad \text{equation satisfied}$$
But the above are just particular solutions. To find all $(x, y)$ pairs, I tried:
$$\begin{aligned}
(x-y)(x^2+xy+y^2) &= 3(x+y)(x-y) \\
x^2+xy+y^2 &= 3(x+y) \end{aligned}\\
\begin{aligned}
x^2+xy+y^2-3x-3y &= 0 \\
x^2+x(y-3)+y(y-3) &=0 \\
 x^2 + (y-3)(x+y) &= 0 \end{aligned}$$
What now? Am I on the right track?

$x=y$ and $x=3$ with $y=0$ (or $x=0$ with $y=3$) are trivial answers. To show there are no others for $x\ne y$, let $n=x+y$, then the expression can be written $x^2+xy+y^2=(x+y)^2 +(3-n)n$ or $x(n-x)+3n-n^2=0$.  Solve the quadratic for $x$ to get $x=\frac{n+\sqrt{12n-3n^2}}{2}$. Since the discriminant has to be real, $n\le 4$, and a perfect square to get an integer solution.  The only possibilities are $n=1$ giving $(2,-1)$ (not allowed) and $n=3$ giving $(3,0)$. 
