Example 4 Given that the three sides of $\triangle A B C$ are $a$, $b$, and $c$, and they satisfy
$$
a b c=2(a-1)(b-1)(c-1) .
$$

Does there exist a $\triangle A B C$ with all sides being integers? If it exists, find the lengths of the three sides; if not, explain the reason.
Let's assume the integers $a \geqslant b \geqslant c$. Clearly, $c \geqslant 2$.
If $c \geqslant 5$, then, $\frac{1}{a} \leqslant \frac{1}{b} \leqslant \frac{1}{c} \leqslant \frac{1}{5}$.
From $a b c=2(a-1)(b-1)(c-1)$, we get
$$
\frac{1}{2}=\left(1-\frac{1}{a}\right)\left(1-\frac{1}{b}\right)\left(1-\frac{1}{c}\right) \geqslant\left(\frac{4}{5}\right)^{3} \text {. }
$$

This is a contradiction.
Therefore, $c$ can only be $2, 3, 4$.
When $c=2$, $a+b=1$, no solution;
When $c=3$, $(a-4)(b-4)=12$, the only solution that can form a triangle is $a=8, b=7, c=3$.

When $c=4$, $(a-3)(b-3)=6$, the only solution that can form a triangle is $a=6, b=5, c=4$.

Therefore, there exists a triangle $\triangle A B C$ with integer side lengths, and the side lengths are $4,5,6$ or $3,7,8$.