Example 4 Find the integer solutions of the equation $x^{2}+x=y^{4}+y^{3}+y^{2}+y$.

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Solution: The original equation can be transformed into: $(2 x+1)^{2}=\left(2 y^{2}+y+1\right)^{2}-y^{2}+2 y$
$$
(2 x+1)^{2}=\left(2 y^{2}+y\right)^{2}+3 y^{2}+4 y+1
$$

From (3) and (4), we know that when $y>2$ or $y<-1$,
$$
\left(2 y^{2}+y\right)^{2}<(2 x+1)^{2}<\left(2 y^{2}+y+1\right)^{2} \text {. }
$$

This equation is obviously not valid (there does not exist a square number between the squares of two consecutive integers). Therefore, we only need to examine the solutions of the equation when $-1 \leqslant y \leqslant 2$.

Thus, we have the solutions for $y=-1,0,1,2$ as $(x, y)=(0,-1)$, $(-1,-1),(0,0),(-1,0),(-6,2),(5,2)$.