Example 5 By Fermat's Little Theorem, for any odd prime $p$, we have $2^{p-1} \equiv 1(\bmod p)$. Question: Does there exist a composite number $n$ such that $2^{n-1} \equiv 1(\bmod n)$ holds?
$$\begin{array}{l}
\text { Hence } \\
\text { Therefore }
\end{array}$$
$$\begin{array}{l}
2^{10}-1=1023=341 \times 3 \\
2^{10} \equiv 1(\bmod 341) \\
2^{340} \equiv 1^{34} \equiv 1(\bmod 341)
\end{array}$$

Hence 341 meets the requirement.
Furthermore, let $a$ be an odd composite number that meets the requirement, then $2^{a}-1$ is also an odd composite number (this can be known through factorization). Let $2^{a-1}-1=a \times q, q$ be an odd positive integer, then
$$\begin{aligned}
2^{2^{a}-1-1}-1 & =2^{2\left(2^{a-1}-1\right)}-1 \\
& =2^{2 a q}-1 \\
& =\left(2^{a}\right)^{2 q}-1 \\
& \equiv 1^{2 q}-1 \\
& \equiv 0\left(\bmod 2^{a}-1\right)
\end{aligned}$$

Therefore $2^{a}-1$ is also a number that meets the requirement. By this recursion (combined with 341 meeting the requirement), it is known that there are infinitely many composite numbers that meet the condition.