To solve this problem, in addition to the exponential properties and other properties given above, the following conclusion is also required: Let $e(k)=3^{k-1}$, then (a) $3^{k} \mid 2^{c(k)}+1$, but $3^{k+1}$ does not divide $2^{c(k)}+1$; (b) $3^{k} \mid 2^{s}+1$ if and only if $3^{k-1} \mid s, s$ is odd. This problem can be considered the most difficult in the IMO, with only one solution method, which seems unsuitable as a competition problem. Below, we only provide the solution steps, and readers are encouraged to complete the rigorous solving process on their own.
(i) There must be $n=3^{a} p_{1}^{a(1)} \cdots p_{s}^{a(s)}, 30$, let $c$ be the smallest positive integer such that $p_{1} \mid 2^{c}+1$, from this we deduce that $c \mid (n, p_{1}-1)$, and thus $c \mid n$, and its prime factors are $<p_{1}$. This implies $c=3^{u}(1 \leqslant u \leqslant a)$. From this, we deduce $a=u=1$, hence $p_{1} \mid 2^{3}+1=9$. This is impossible.
(iv) Only $n=1,3$. Therefore, $n=3$.