$12 \cdot 80$ Find all integer solutions $x$ and $y$ to the equation $x^{2}=1+4 y^{3}(y+2)$. (Canadian Mathematical Olympiad Training Problem, 1992)
[Solution] (1) If $y=0$ or $y=-2$, then $x^{2}=1, x= \pm 1$.
(2) If $y=-1$, then $x^{2}=-3$, no integer solution.
(3) If $y \geqslant 1$, then $4 y^{2}>1>1-4 y$.
$$
\begin{aligned}
\left(2 y^{2}+2 y\right)^{2} & =4 y^{4}+8 y^{3}+4 y^{2} \\
& >4 y^{4}+8 y^{3}+1>4 y^{4}+8 y^{3}-4 y+1 \\
& =\left(2 y^{2}+2 y-1\right)^{2}
\end{aligned}
$$

Thus, $\left(2 y^{2}+2 y-1\right)^{2}>4[1-y(y+2)]$,
$$
\begin{aligned}
\left(2 y^{2}+2 y-1\right)^{2} & =4 y^{4}+8 y^{3}-4 y+1>4 y^{4}+8 y^{3}+1 \\
& >4 y^{4}+8 y^{3}-4 y^{2}-8 y+4 \\
& =\left(2 y^{2}+2 y-2\right)^{2},
\end{aligned}
$$

which means $\left(2 y^{2}+2 y-2\right)^{2}<x^{2}<\left(2 y^{2}+2 y-1\right)^{2}$.
Therefore, $x$ is not an integer.
From (1), (2), (3), and (4), we can conclude that the equation has only four solutions:
$$
(x, y)=(1,0),(1,-2),(-1,0),(-1,-2) .
$$