Example 4 Find all positive integers $n$, such that $2^{n} \equiv 1(\bmod n)$.

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When $n=1$, it is obviously true.
Below is the proof: When $n>1$, we always have
$$2^{n} \not \equiv 1(\bmod n)$$

In fact, if there exists $n>1$, such that
$$2^{n} \equiv 1(\bmod n),$$

take the smallest prime factor $p$ of $n$, then $2^{n} \equiv 1(\bmod p)$. By Fermat's Little Theorem, we know $2^{p-1} \equiv$ $1(\bmod p)$, so,
$$2^{(n, p-1)} \equiv 1(\bmod p)$$

Using the fact that $p$ is the smallest prime factor of $n$, we know $(n, p-1)=1$, leading to $2 \equiv 1(\bmod$ $p)$, i.e., $p \mid 1$. This is a contradiction. $\square$

In summary, only $n=1$ meets the requirement.