Example 10 Try to find all positive integers $a, b, c$, where $1<a<b<c$, and such that $(a-1)(b-1)(c-1)$ is a divisor of $a b c-1$.

Find all positive integers $a, b, c$, where $1<a<b<c$, and such that $(a-1)(b-1)(c-1)$ is a divisor of $a b c-1$.
Analysis and Solution According to the problem, there exists $k \in \mathbf{N}$, such that
$$
a b c-1=k(a-1)(b-1)(c-1)
$$

By the discreteness of integers, we have $a \leqslant b-1, a \leqslant c-2$,
$$
\begin{array}{l}
\text { Hence }(k-1)(a-1)(b-1)(c-1)=(a b c-1)-(a-1)(b-1)(c-1) \\
\quad=a b+b c+c a-a-b-c=a(b-1)+b(c-1)+c(a-1) \\
\quad \leqslant(c-2)(b-1)+b(c-1)+c(b-2)=3(b c-b-c+1)-1 \\
\quad0$, so $k \neq 1$, hence $(k-1) \cdot(a-1)=1$ or $(k-1)(a-1)=2$.
(1) If $(k-1)(a-1)=1$ holds, then $k=a=2$. At this time, the left side of (1) is odd, and the right side is even, which is a contradiction.
(2) If $(k-1)(a-1)=2$ holds, then $a=3, k=2$

or $a=2, k=3$
Substituting (2) into (1), we get $3 b c-1=4(b-1)(c-1)=4(b c-b-c+1)$, i.e., $b c-4(b+c)+16=11$, i.e., $(b-4)(c-4)=11$.
Also, since $c>b$, we have $b-4=1, c-4=11$,
so $b=5, c=15$.
Similarly, substituting (3) into (1) yields $b=4, c=8$.
Upon verification, we get $a=3, b=5, c=15$, or $a=2, b=4, c=8$.