Example 3 Find all triples $(x, y, n), x, y, n \in \mathbf{N}^{*}$ satisfying $\frac{x!+y!}{n!}=3^{n}$.
(2005 Vietnam Mathematical Olympiad)
Analysis From the $3^{n}$ on the right, we can preliminarily estimate that $n$ cannot be too large, so we need to further determine the range of $n$.
Solution (1) is
$$
x!+y!=3^{n} \cdot n!.
$$

Assume without loss of generality that $x \geqslant y$. The power of the factor 3 in $y!$ is
$$
3(y!)=\left[\frac{y}{3}\right]+\left[\frac{y}{3^{2}}\right]+\cdots+\left[\frac{y}{3^{n}}\right]+\cdots
$$
The power of the factor 3 in $x!$ is
$$
3(x!)=\left[\frac{x}{3}\right]+\left[\frac{x}{3^{2}}\right]+\cdots+\left[\frac{x}{3^{n}}\right]+\cdots
$$

Since $x \geqslant y$, we have $3(x!) \geqslant 3(y!)$. Therefore, the highest power of 3 on the left side of (2) is $3(y!)$, so $3(y!) \geqslant n$.
Also, $\left[\frac{y}{3}\right]+\left[\frac{y}{3^{2}}\right]+\cdots+\left[\frac{y}{3^{n}}\right]+\cdots \leqslant \frac{y}{3}+\frac{y}{3^{2}}+\cdots+\frac{y}{3^{n}}+\cdots n$, so $y>2 n$.
Then the left side of (2) $>(2 n)!+(2 n)!=n!\cdot 2 \cdot(2 n)(2 n-1) \cdots(n+1)$.
If $n \geqslant 2$, then the left side $>n!\cdot 2 \cdot 3^{n}>$ the right side, no solution. Therefore, $n=1, x!+y!=3$; so $x=1, y=2$ or $x=2, y=1$.
In summary, the solutions are $(x, y, n)=(2,1,1),(1,2,1)$.
/Comment According to the characteristics of the problem, we examined the prime factors on both sides, where the $p$ component is an important tool.