Example 5 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)
- Analysis: First, estimate
$$
s=\frac{a b c-1}{(a-1)(b-1)(c-1)}(s \in \mathbf{N})
$$

the range. For this, let $c=a-1, y=b-1, z=c-1$, then $1 \leqslant x1$;
$$
\text { Also, } \begin{aligned}
s & =\frac{(x+1)(y+1)(z+1)-1}{x y z} \\
& <\frac{(x+1)(y+1)(z+1)}{x y z} \\
& =\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right) \\
& \leqslant\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)=4,
\end{aligned}
$$
$\therefore 1<s<4$, i.e., $s=2$ or $s=3$.
(1) If $s=2$, then
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}=1 \text {. }
$$

Solving gives $(x, y, z)=(2,4,14)$.
(2) If $s=3$, then
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}=2 \text {. }
$$

Solving gives $(x, y, z)=(1,3,7)$.
In summary, the positive integer solutions that satisfy the problem are $(3,5,15)$ and $(2,4,8)$.