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

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Solution: Multiply by 4 to complete the square, getting $(2 x+1)^{2}$
$$
\begin{array}{l}
=4\left(y^{4}+y^{3}+y^{2}+y\right)+1 \\
=\left(2 y^{2}+y+1\right)^{2}-y^{2}+2 y \\
=\left(2 y^{2}+y\right)^{2}+3 y^{2}+4 y+1 .
\end{array}
$$
$\therefore$ When $y>2$ or $y<-1$, $\left(2 y^{2}+y\right)^{2}<(2 x+1)^{2}<\left(2 y^{2}+y+1\right)^{2}$,
$\therefore-1 \leqslant y \leqslant 2, y=-1,0,1,2$, the solutions to the equation are $(x, y)=(0,-1),(-1,-1),(0,0),(-1,0),(-6,2),(5,2)$.