Example 6 Find all positive integers $a, b, c$ such that $1 < a < b < c$ and $(a-1)(b-1)(c-1)$ is a divisor of $abc-1$.
(33rd IMO)
First, estimate
$$
s=\frac{a b c-1}{(a-1)(b-1)(c-1)}\left({ }_{s} \in \mathbf{N}_{+}\right)
$$

the range.
Let $x=a-1, y=b-1, z=c-1$, then $1 \leqslant x$ $1, \\
s<\frac{(x+1)(y+1)(z+1)}{x y z} \\
=\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right) \\
\leqslant\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)=4,
\end{array}
$$

Therefore, $s=2$ or 3.
(1) If $s=2$, then
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}=1 .
$$

Clearly, $x \neq 1$. If $x \geqslant 3$, then $y \geqslant 4, z \geqslant 5$. Hence
$$
\begin{array}{l}
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x} \\
\leqslant \frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{12}+\frac{1}{20}+\frac{1}{15}=\frac{59}{60}<1,
\end{array}
$$

which contradicts equation (1).
Therefore, $x=2$.
In this case, $\frac{1}{y}+\frac{1}{z}+\frac{1}{2 y}+\frac{1}{y z}+\frac{1}{2 z}=\frac{1}{2}$.
It is easy to see that $3 \leqslant y \leqslant 5$ (when $y \geqslant 6$, $z \geqslant 7$, the above equation does not hold). Solving, we get $(x, y, z)=(2,4,14)$.
(2) If $s=3$, then
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}=2 \text{. }
$$

Thus, $x=1$ (otherwise, the left side of equation (2) $\leqslant \frac{1}{2}+\frac{1}{3}+$ $\left.\frac{1}{4}+\frac{1}{6}+\frac{1}{12}+\frac{1}{8}<2\right)$, equation (2) becomes
$$
\frac{2}{y}+\frac{2}{z}+\frac{1}{y z}=1 \text{. }
$$

Similarly, $2 \leqslant y \leqslant 3$.
Solving, we get $(x, y, z)=(1,3,7)$.
In summary, the positive integer solutions that satisfy the problem are
$$
(3,5,15) \text{ and }(2,4,8) \text{. }
$$