The relation in the problem can be written in the form
$$x! + y! = 3^n n! \quad (1)$$
Suppose that $(x, y, n)$ is a triple of natural numbers satisfying (1).
It is easily seen that $n \ge 1$ and w.l.g. we can suppose that $x \le y$. We must now consider the following cases.

1) 1st case: $x \le n$
$$\text{It is clear that } (1) \Leftrightarrow 1 + \frac{y!}{x!} = 3^n \frac{n!}{x!}. \qquad (2)$$
(2) implies that $1 + \frac{y!}{x!} \equiv 0 \pmod{3}$. Since the product of three consecutive integers is divisible by 3 and since $n \ge 1$, we have: $x < y \le x + 2$.

a) If $y = x + 2$, (2) implies
$$1 + (x + 1)(x + 2) = 3^n \frac{n!}{x!} \qquad (3)$$
Since the product of two consecutive integers is divisible by 2, from (3), we deduce that $n \le x + 1$.
- If $n = x$, (3) implies
$$1 + (x + 1)(x + 2) = 3^x, \quad \text{i.e. } x^2 + 3x + 3 = 3^x \qquad (4)$$
Since $x \ge 1$, (4) shows that $x \equiv 0 \pmod{3}$, therefore $x \ge 3$ and we get a contradiction
$$3 = x^2 + 3x - 3^x \equiv 0 \pmod{9}$$
which proves that $n \ne x$.
- If $n = x + 1$, (3) implies that
$$1 + (x + 1)(x + 2) = 3^n(x + 1)$$
hence $x + 1$ is a positive divisor of 1. Therefore $x = 0$ and consequently $y = 2, n = 1$.

b) If $y = x + 1$, (2) implies
$$x + 2 = 3^n \frac{n!}{x!} \qquad (5)$$
Since $x \ge 1$, (5) shows that $x \ge 1$ and then $x = n$ and
$$x + 2 = 3^x. \qquad (6)$$
$x = 1$ is the unique natural number satisfying (6) therefore, in this case, if the triples $(x, y, n)$ satisfy (1) then $(x, y, n) = (0, 2, 1)$ or $(x, y, n) = (1, 2, 1)$.

2) 2nd case: $x > n$.
It is clear that
$$(1) \Leftrightarrow \frac{x!}{n!} + \frac{y!}{n!} = 3^n. \qquad (7)$$
Since $n + 1$ and $n + 2$ can not be simultaneously the powers of 3, from (7), we deduce that $x = n + 1$. Then (2) implies that
$$n + 1 + \frac{y!}{n!} = 3^n. \qquad (8)$$
Since $y \ge x$, we see that $y \ge n + 1$. By putting $A = \frac{y!}{(n+1)!}$, we can write (8) in the form
$$(n + 1)(1 + A) = 3^n. \qquad (9)$$
It is clear that if $y \ge n + 4$ then $A \equiv 0 \mod 3$, and $A + 1$ can not be a power of 3. Hence (9) shows that $y \le n + 3$.
So $n + 1 \le y \le n + 3$.

a) If $y = n + 3$ then $A = (n + 2)(n + 3)$, and from (9) we get
$$(n + 1)(1 + (n + 2)(n + 3)) = 3^n \text{ i.e. } (n + 2)^3 - 1 = 3^n. \qquad (10)$$
It follows that $n > 2$ and $n + 2 = 1 \mod 3$. By putting $n + 2 = 3k + 1, k > 2$, we can write (10) in the form
$$9k(3k^2 + 3k + 1) = 3^{3k-1}.$$
It implies that $3k^2 + 3k + 1$ is a power of 3. This contradiction proves that $y \ne n + 3$.

b) If $y = n + 2$ then $A = n + 2$ and from (9) we get
$$(n + 1)(n + 3) = 3^n. \qquad (11)$$

c) If $y = n + 1$ then $A = 1$ and from (9) we get
$$2(n + 1) = 3^n.$$
It is clear that there exist no $n$ satisfying this relation. So $y \ne n + 1$.

Thus, if the triple $(x, y, n)$, with $x \le y$, satisfies (1) then $n \ge x$.
Consequently, if $(x, y, n)$ is a triple of natural numbers satisfying (1) then $(x, y, n) = (0, 2, 1)$ or $(x, y, n) = (2, 0, 1)$, or $(x, y, n) = (2, 1, 0)$, or $(x, y, n) = (2, 1, 1)$.
By direct verification, we see that the four mentioned triples satisfy (1). So these triples are all triples satisfying the conditions of the problem.