Example 8 Find all integers $a, b, c$, where $1<a<b<c$, such that $(a-1)(b-1)(c-1)$ is a divisor of $a b c-1$.

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To solve this problem, we first need to estimate the range of values for \( s = \frac{a b c - 1}{(a-1)(b-1)(c-1)} \left(s \in \mathbf{N}^{*}\right) \), and then check each possible case to determine the solution.

Solution: Given \( a \geqslant 2, b \geqslant 3, c \geqslant 4 \), we have:
\[
\frac{a-1}{a} = 1 - \frac{1}{a} \geqslant \frac{1}{2}, \quad \frac{b-1}{b} \geqslant \frac{2}{3}, \quad \frac{c-1}{c} \geqslant \frac{3}{4}.
\]

Thus,
\[
\frac{(a-1)(b-1)(c-1)}{a b c} \geqslant \frac{1}{4},
\]
which implies
\[
a b c \leqslant 4(a-1)(b-1)(c-1).
\]
Therefore,
\[
s = \frac{a b c - 1}{(a-1)(b-1)(c-1)} > 1.
\]
From \( a > 1 \), we get \( a \geqslant 3, b \geqslant 5, c \geqslant 7 \).
If \( b \geqslant 7 \), then \( c \geqslant 9 \), and thus
\[
\frac{(a-1)(b-1)(c-1)}{a b c} = \left(1 - \frac{1}{a}\right)\left(1 - \frac{1}{b}\right)\left(1 - \frac{1}{c}\right) \geqslant \frac{2}{3} \cdot \frac{6}{7} \cdot \frac{8}{9} = \frac{32}{63} > \frac{1}{2}.
\]

This contradicts (1). Therefore, we must have \( b = 5 \) and \( a = 3 \). Substituting into (1), we get \( 16(c-1) = 15c - 1 \), which simplifies to \( c = 15 \).

(3) If \( s = 3 \), then \( 3(a-1)(b-1)(c-1) = a b c - 1 \frac{1}{3} \).

This contradicts (2). Therefore, we must have \( a = 2 \).
If \( b \geqslant 5 \), then
\[
\frac{(a-1)(b-1)(c-1)}{a b c} \geqslant \frac{1}{2} \cdot \frac{4}{5} \cdot \frac{5}{6} = \frac{1}{3},
\]
which contradicts (2). Therefore, we must have \( b = 3 \) or \( b = 4 \).
If \( b = 3 \), from (2) we get \( 6(c-1) = 6c - 1 \), which has no solution;
If \( b = 4 \), from (2) we get \( 9(c-1) = 8c - 1 \), which simplifies to \( c = 8 \).

In summary, the integer solutions \((a, b, c)\) that satisfy the conditions are \((3, 5, 15)\) and \((2, 4, 8)\).