Example I Solve the equation
$\left(x_{1}^{2}+1\right)\left(x_{2}^{2}+2^{2}\right) \cdots\left(x_{n}^{2}+n^{2}\right)=2^{n} \cdot n!x_{1} \cdot x_{2} \cdots x_{n}$, where $n \in \mathbf{N}^{*}, x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}$.
Given that the equation has $n$ unknowns, applying the general method to solve the equation is very difficult. Considering $x_{1}^{2}+1 \geqslant 2 x_{1}, x_{2}^{2}+2^{2} \geqslant 2 \cdot 2 x_{2}, \cdots, x_{n}^{2}+n^{2} \geqslant 2 n x_{n}$, try to use the inequality relationship to help solve it.
Solution: Since $x_{1}^{2}+1 \geqslant 2 x_{1}$,
$$
\begin{array}{l}
x_{2}^{2}+2^{2} \geqslant 2 \cdot 2 x_{2}, \\
\cdots \cdots \\
x_{n}^{2}+n^{2} \geqslant 2 n \cdot x_{n} .
\end{array}
$$

Therefore, $\left(x_{1}^{2}+1\right)\left(x_{2}^{2}+2^{2}\right) \cdots\left(x_{n}^{2}+n^{2}\right) \geqslant 2^{n} \cdot n!x_{1} \cdot x_{2} \cdots x_{n}$.
The equality holds if and only if $x_{1}=1, x_{2}=2, \cdots, x_{n}=n$, hence the solution to the original equation is $x_{1}=1, x_{2}=2, \cdots$,
$$
x_{n}=n
$$