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

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Solving as in the previous example, multiply both sides of the equation by 4 and complete the square on the left side:
$$(2 x+1)^{2}=4\left(y^{4}+y^{3}+y^{2}+y\right)+1$$

Next, we estimate the right side of equation (1). Since
$$\begin{aligned}
4\left(y^{4}+y^{3}+y^{2}+y\right)+1 & =\left(2 y^{2}+y+1\right)^{2}-y^{2}+2 y \\
& =\left(2 y^{2}+y\right)^{2}+3 y^{2}+4 y+1
\end{aligned}$$

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 $2 y^{2}+y$ and $2 y^{2}+y+1$ are two consecutive integers, there cannot be a perfect square between their squares, so the above inequality does not hold.

Therefore, we only need to consider the solutions of the equation when $-1 \leqslant y \leqslant 2$, which is trivial. It is easy to find that all integer solutions to the original equation are
$$(x, y)=(0,-1),(-1,-1),(0,0),(-1,0),(-6,2),(5,2) .$$