$(x+1)^2 + (y+1)^2 + xy(x+y+3)=2$ I've came across this problem some hours ago and, although it looks (and possibly is) just some algebra calculus, I can't get on the right track.

Find $x$, $y$ integers such that
  $$ (x+1)^2 + (y+1)^2 + xy(x+y+3)=2 $$

Some basic algebra will lead to:
$$
(x+y)^2 + x(xy+2) + y(xy+2) + xy = 0$$
 $$(x+y)(xy+x+y+2)=-xy$$
Now, some divisibility properties should end the problem.
Well, maybe my approach isn't the right one, but I'm just stuck at this point.
A piece of advice or a hint would be apreciated.

The origin $(0,0)$, $(0,-2)$ and $(-2,0)$ are the only diophantine (integer) solutions. Simplifying the equation and taking note of the symmetry about $y=x$ ($x$ & $y$ are interchangeable, so we will eventually only see symmetric terms such as xy and x+y), we see that
$$\begin{align}
0 &= (x+1)^2 + (y+1)^2 + xy(x+y+3) - 2
\\&= x^2+2x + y^2+2y + xy(x+y+3)
\\&= x^2+y^2 + 2(x+y) + xy(x+y+3)
\\&= (x+y)^2 + 2(x+y) + xy(x+y+1)
\\&= (x+y+1)^2 + xy(x+y+1) - 1
\\&= u^2 + 2uv - 1
\\&= \left(u+v\right)^2 - \left(1+v^2\right)
\end{align}$$
for $u=x+y+1$ & $v=xy/2$, a quadratic equation with solution $u=-v\pm\sqrt{1+v^2}$, or $2(x+y+1)=-xy\pm\sqrt{4+(xy)^2}$. Next, note that $x$ & $y$ can't both be odd, because if they were, then we would have $u$ odd $\implies 2v=1-u^2$ even $\implies v\in\mathbb{Z}\implies xy$ even, leading to a contradiction. But then $2|xy$ so that $v\in\mathbb{Z}$. However, then the radical $\sqrt{1+v^2}$ can only be an integer if $v^2$ is both a perfect square and one less than a perfect square. This can only happen when $v=xy=0$, i.e., when at least one of $x,y$ is zero as noted above.
Another approach:
$$\begin{align}0
&=(x+y)^2+x(xy+2)+y(xy+2)+xy\\
&=(x+y)^2+(x+y)(xy+2)+xy\\
&=u^2+ut+t-2
\end{align}$$
for (integers) $$\begin{align}t&=xy+2\\u&=x+y\end{align}$$ so that
$$t(1+u)=2-u^2$$ or $$t=\frac{2-u^2}{1+u}=\frac{1-u}{1+u}+\frac1{1+u}.$$
For $t$ to be an integer, we must have $u+1=\pm1$, or $u=x+y\in\{0,-2\}$.
In both cases, $t=2\implies xy=0\implies x$ or $y$ is zero.
This only admits the three solutions $(x,y)=(0,0),(0,-2),(-2,0)$.
