Find all integer solutions of $(a+b+3)^2+2ab=3ab(a+2)(b+2)$ Here is another number theory problem that I am not able to do
Find all integer solutions of $$(a+b+3)^2+2ab=3ab(a+2)(b+2)$$
My attempt$:$
On expanding we get,
$$b^2+4ab+6b+a^2+6a+9=3a^2b^2+6a^2b+6ab^2+12ab$$ or $$-6ab^2+b^2-8ab+6b+a^2+6a+9=3a^2b^2+6a^2b$$ or $$-6ab^2+b^2-8ab+6b+a^2+6a+9-3a^2b^2-6a^2b=0$$ or $$\left(1-3b^2-6b\right)a^2+\left(-6b^2-8b+6\right)a+b^2+6b+9=0$$ or $$a=\frac{-\left(-6b^2-8b+6\right)\pm \sqrt{\left(-6b^2-8b+6\right)^2-4\left(1-3b^2-6b\right)\left(b^2+6b+9\right)}}{2\left(1-3b^2-6b\right)}$$ or $$a=\frac{-\left(-6b^2-8b+6\right)\pm \:4\sqrt{3}\left(b+1\right)\sqrt{b\left(b+2\right)}}{2\left(1-3b^2-6b\right)};\quad \:b\ne \:-\frac{3+2\sqrt{3}}{3},\:b\ne \frac{2\sqrt{3}-3}{3}$$ and after further solving $$a=\frac{3b^2+4b-3+2\sqrt{3}b\sqrt{b\left(b+2\right)}+2\sqrt{3}\sqrt{b\left(b+2\right)}}{1-3b^2-6b}$$ and $$a=\frac{3b^2+4b-3-2\sqrt{3}b\sqrt{b\left(b+2\right)}-2\sqrt{3}\sqrt{b\left(b+2\right)}}{1-3b^2-6b}$$ where $b\ne \:-\frac{3+2\sqrt{3}}{3},\:b\ne \frac{2\sqrt{3}-3}{3}$
So, now I can assign some integer values to $a$ and check if that value of $a$ is giving some integer values of $b$ or not. Like if $a=0$ then $b=-3$. But I think it is merely hit and trial because this thing can also apply to the original equation. Any help will be greatly appreciated.

Let x=a+1, y = b+1, so
$(x+y+1)^2 + 2(x-1)(y-1) = 3(x^2-1)(y^2-1)$.
The right hand side is quite obviously larger when |x|, |y| >= 3. Finding solutions for x=-2, -1, 0, 1, 2 are easy to find.
