Problem 1597

[Proof] Let $x, y, z$ be positive integer solutions of the original equation that meet the problem's requirements. Then $x, y, z$ are pairwise distinct, and there exist positive integers $u, v, w$ such that
$$\left\{\begin{array}{l} x y=z u \\ y z=x v \\ z x=y w . \end{array}\right.$$
(1) $\times$ (2) $\times$ (3) gives $(x y z)^{2}=(x y z)(u v w)$,

Noting that $x y z \neq 0$, we have
$$x y z=\text { uvw. } .$$

Substituting (1), (2), and (3) into the left side of (4), we get
$$z^{2} u=x^{2} v=y^{2} w=x y z=u v w$$

Thus, we have

i.e., $\square$
$$x^{2}=u w, y^{2}=u v, z^{2}=v w,$$

Let
$$\begin{array}{l} x=\sqrt{u w}, y=\sqrt{u v}, z=\sqrt{v w} . \\ n=\sqrt{u}, m=\sqrt{w}, k=\sqrt{v} \end{array}$$

Then $\square$
$$x=n m, y=n k, z=m k \text {. }$$

Substituting (5) into the original equation, we get
$$n m-n k+m k=1$$

i.e.,
$$n(k-m)=m k-1$$

Taking $k-m=1$, then
$$k=m+1, \quad n=m^{2}+m-1$$

Substituting (6) into (5) gives
$$x=m\left(m^{2}+m-1\right), y=(m+1)\left(m^{2}+m-1\right), z=m(m+1).$$
Upon verification, when $m=2,3,4, \cdots \cdots$, the above expressions provide infinitely many sets of positive integer solutions that meet the requirements of the original equation.