Example 8 Try to find the integer solutions of the equation $x^{2}+x=y^{4}+y^{3}+y^{2}+y$.
Multiply both sides of the equation by 4, and complete the square on the left side, we get
$$
(2 x+1)^{2}=4\left(y^{4}+y^{3}+y^{2}+y\right)+1 \text {. }
$$

The right side of (1) $=\left(2 y^{2}+y+1\right)^{2}-y^{2}+2 y=\left(2 y^{2}+y\right)^{2}+3 y^{2}+4 y+1$.
Thus, when $y>2$ or $y<-1$, we have $\left(2 y^{2}+y\right)^{2}<(2 x+1)^{2}<\left(2 y^{2}+y+1\right)^{2}$.
Since $2 y^{2}+y, 2 y^{2}+y+1$ are two consecutive integers, there cannot be a perfect square between them, so the above inequality does not hold, hence $-1 \leqslant y \leqslant 2$. Therefore, the complete set of integer solutions to the original equation is
$$
(x, y)=(0,-1),(-1,-1),(0,0),(-1,0),(-6,2),(5,2) .
$$