Problem 779

Notice,
$$\begin{array}{l} (x y-1)(y z-1)(z x-1) \\ =x^{2} y^{2} z^{2}-(x+y+z) x y z+x y+y z+z x-1 . \end{array}$$

The problem is equivalent to finding distinct positive integers $x, y, z$ such that $x y z \mid (x y + y z + z x - 1)$.
Assume without loss of generality that $x > y > z$.
Then $x y z \leq x y + y z + z x - 1$. If $z = 1$, then $x y \leq x y + y + x - 1$, which is a contradiction since $y$ is a positive integer.
$$\begin{array}{l} \text { (2) When } z=2 \text {, } \\ 2 x y \mid(x y+2 y+2 x-1) \\ \Rightarrow 2 x y \leq x y+2 y+2 x-1 \\ \Rightarrow x y \leq 2 y + 2 x - 1. \end{array}$$
Since $z=2$, we know $y=3$.
Thus, $6 x \mid (5 x + 5) \Rightarrow 6 x \leq 5 x + 5 \Rightarrow x \leq 5$.
Combining $x > y = 3$, we know $x = 4, 5$.
Upon inspection, $x = 5$ satisfies the condition.
Therefore, the positive integer solutions $(x, y, z)$ that satisfy the condition are $(2, 3, 5)$ and their permutations.