Consider the sequence $a_1, a_2, \dots$ defined by $a_1 = 2$ and $a_{n+1} = 2^{a_n} + 2$ for all $n$. It is easy to show that $\tfrac{a_{n+1}}{a_n}$ and $\tfrac{a_{n+1} - 1}{a_n - 1}$ are odd integers for all $n$, so it follows that the set \[\left\{2^{a_n} - 2 : n = 1, 2, \dots\right\}\]works. (If $\tfrac{n}{m}$ and $\tfrac{n-1}{m-1}$ are odd integers then $2^m + a \mid 2^n + a$ for all $a \in \{-2, \dots, 2\}$.)