Example 14 Solve the equation
\[
\begin{array}{l}
(x+1)(x+2)(x+3)(x+4) \\
=(x+1)^{2}+(x+2)^{2}+(x+3)^{2}+(x+4)^{2} .
\end{array}
\]
Solution: Let $y=\frac{1}{4}(x+1+x+2+x+3+x+4)=x+\frac{5}{2}$, the original equation can be transformed into
$$
\begin{array}{l}
\left(y-\frac{3}{2}\right)\left(y-\frac{1}{2}\right)\left(y+\frac{1}{2}\right)\left(y+\frac{3}{2}\right) \\
=\left(y-\frac{3}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2} \\
\quad+\left(y+\frac{3}{2}\right)^{2} .
\end{array}
$$

Rearranging, we get $16 y^{4}-104 y^{2}-71=0$.
Solving, we get $y^{2}=\frac{13+4 \sqrt{15}}{4}$
or $y^{2}=\frac{13-4 \sqrt{15}}{4}<0$ (discard).
$\therefore \quad y= \pm \frac{1}{2} \sqrt{13+4 \sqrt{15}}$.