Example 40 (1992 Canadian Mathematical Olympiad Training Problem) Find all integer solutions $x$ and $y$ that satisfy the equation $x^{2}=1+4 y^{3}(y+2)$.
(1) If $y=0$ or $y=-2$, then $x^{2}=1, x= \pm 1$.
(2) If $y=-1$, then $x^{2}=-3$, no integer solutions.
(3) If $y \geqslant 1$, then $4 y^{2}>1>1-4 y$.
$$
x^{2}=4 y^{4}+8 y^{3}+14 y^{4}+8 y^{3}-4 y+1=\left(2 y^{2}+2 y-1\right)^{2},
$$
which means $x^{2}$ lies between the squares of two consecutive integers, so $x$ cannot be an integer.
(4) If $y \leqslant-3$, then $1-4 y>1>4[1-y(y+2)]$.
$$
\begin{array}{l}
x^{2}=4 y^{4}+8 y^{3}+14 y^{4}+8 y^{3}+4[1-y(y+2)]=\left(2 y^{2}+2 y-2\right)^{2},
\end{array}
$$

Similarly, $x$ cannot be an integer.
From (1), (2), (3), (4), we know that the equation has only 4 solutions:
$$
(x, y)=(1,0),(1,-2),(-1,0),(-1,-2) .
$$