Prove (i) For any odd number $a$, there must be a $d \geqslant 2$ such that $a \mid 2^{d}-1$, and $a$ and $2^{d}-3$ are coprime (Chapter 1, §3, Example 5).
(ii) Based on (i), use induction to construct the sequence. Take $a_{1}=2^{2}-3$. If $a_{1}, \cdots, a_{k}$ have been chosen, then take
$a_{k+1}=2^{d(k+1)}-3$, where $d(k+1)$ satisfies $a_{1} \cdots a_{k} \mid 2^{d(k+1)}-1$. Thus, an infinite subsequence $a_{1}, \cdots, a_{k}, \cdots$ of the sequence $2^{n}-3(n=2,3, \cdots)$ is inductively defined.
(iii) From $a_{1} \cdots a_{k} \mid a_{k+1}+2$ and the fact that $a_{j}$ are all odd, the desired conclusion follows (why).