Example 3 Find all triples $(x, y, n), x, y, n \in \mathbf{N}^{\cdot}$ satisfying $\frac{x!+y!}{n!}=3^{n}$.

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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
$$
x! + y! = 3^{n} \cdot n! \text{.}
$$

Assume without loss of generality that $x \geqslant y$. The exponent 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 exponent 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 $y > 2n$.
Thus, the left side of (2) $>(2n)! + (2n)! = n! \cdot 2 \cdot (2n)(2n-1) \cdots (n+1)$.
If $n \geqslant 2$, then the left side $> n! \cdot 2 \cdot 3^{n} >$ the right side, so there is no solution. Therefore, $n = 1$, and $x! + y! = 3$; so $x = 1, y = 2$ or $x = 2, y = 1$.
In summary, the solutions are $(x, y, n) = (2, 1, 1), (1, 2, 1)$.