Solve $x^2y^2 - 4x^2y + y^3 + 4x^2 - 3y^2 + 1 = 0$ over the integers. 
Solve $$x^2y^2 - 4x^2y + y^3 + 4x^2 - 3y^2 + 1 = 0$$ over the integers.

You can probably guess by now... This problem is adapted from a recent competition.
If there are any other solutions, please post them below. I have provided one if you want to check out.

Not suprisingly,
$$x^2y^2 - 4x^2y + y^3 + 4x^2 - 3y^2 + 1 = 0$$
$$\iff (x^2y^2 + y^3 + y^2) - (4x^2y + 4y^2 + 4y) + (4x^2 + 4y + 4) = 3$$
$$\iff y^2(x^2 + y + 1) - 4y(x^2 + y + 1) + 4(x^2 + y + 1) = 3$$
$$\iff (y^2 - 4y + 4)(x^2 + y + 1) = 3 \iff (y - 2)^2(x^2 + y + 1) = 3$$
$$\implies (y - 2)^2 \mid 3 \implies (y - 2)^2 \in \{\pm 1, \pm 3\}$$
Having said that, $(y - 2)^2$ is a perfect square. $\implies (y - 2)^2 = 1 \implies \left\{ \begin{align} x^2 + y + 1 = 3\\ y - 2 = \pm 1 \end{align} \right.$.
$$\implies x^2 = -(y - 2) = \mp 1 \implies x^2 = -(y - 2) = 1 \iff \left\{\begin{align} x = \pm 1\\ y = 1\end{align} \right.$$
