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 $y>x$, then $c>b$, and
$$\frac{c}{b}=\frac{y}{x}=\frac{a^{c}}{a^{b}}=a^{a^{c-b}},$$

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, it must be that $x=2, k=2$. 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.