Example 9 Find all positive integers $a, b, c$, where $1<a<b<c$, such that $(a-1)(b-1)(c-1)$ is a divisor of $abc-1$.

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Let $x=a-1, y=b-1, z=c-1$, then $1 \leqslant x<y<z$. Also, because
$$
\begin{aligned}
z & \leqslant \frac{(x+1)(y+1)(z+1)-1}{x y z}<\frac{(x+1)(y+1)(z+1)}{x y z} \\
& =\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right) \leqslant\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)=4,
\end{aligned}
$$

so $\frac{(x+1)(y+1)(z+1)-1}{x y z}=2$ or 3.
(1) If the above is 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 $y \geqslant 4, z \geqslant 5$, thus
$$
\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x} \leqslant \frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{12}+\frac{1}{20}+\frac{1}{15}=\frac{59}{60}<1 \text {, }
$$

contradicting (1). Therefore, $x=2$, at this point,
$$
\frac{1}{y}+\frac{1}{z}+\frac{1}{2 y}+\frac{1}{y z}+\frac{1}{2 z}=\frac{1}{2} \text {. }
$$

It is easy to see that $3 \leqslant y \leqslant 5$ (when $y \geqslant 6$, $z \geqslant 7$, the above does not hold), only when $y=4$, $z=14$ is an integer, hence (1) has only one set of positive integer solutions $(2,4,14)$.
(2) If $\frac{(x+1)(y+1)(z+1)}{x y z}=3$, 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 (2) $\left.\leqslant \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{12}+\frac{1}{8}<2\right)$, (2) becomes
$$
\frac{2}{y}+\frac{2}{z}+\frac{1}{y z}=1
$$

It is easy to see that $2 \leqslant y \leqslant 3$, only when $y=3$, $z=7$ is an integer.
Hence, the positive integer solutions satisfying (2) are $(x, y, z)=(1,3,7)$.
In summary, the positive integers $(a, b, c)$ that satisfy the problem are $(3,5,15),(2,4,8)$.