4・169 Solve the equation in the set of natural numbers
$$
x^{y}=y^{x}(x \neq y) .
$$
[Solution] If $x=1$, then $y=1$, which contradicts $x \neq y$. Therefore, $x \geqslant 2$ and $y \geqslant 2$.

Let $x^{y}=y^{x}=a^{d}$, where $a, d$ are natural numbers, and make $d$ as large as possible. Then $a \neq 1$, i.e., $a \geqslant 2$, and there must exist $b, c$ such that
$$
x=a^{b}, y=a^{c}.
$$

Substituting into the original equation, we get
$$
a^{b y}=a^{c x}.
$$

Thus, $b y=c x$.
Assume without loss of generality that $y>x$, so $c>b$, and
$$
\frac{c}{b}=\frac{y}{x}=\frac{a^{c}}{a^{b}}=a^{c-b},
$$

Let $a^{c-b}=k$, then $y=k x(k \geqslant 2)$. Substituting into the original equation, we get
$$
x^{k x}=(k x)^{x},
$$

Thus,
$$
x^{k}=k x,
$$

i.e., $x^{k-1}=k \quad(x \geqslant 2, k \geqslant 2)$.
Therefore, we must have $x=2, k=2$. Hence, $y=k x=4$. Upon verification, $x=2, y=4$ is a solution to the original equation. By symmetry, $x=4, y=2$ is also a solution to the original equation.