Can it be shown that for any two arbitrary sequences of the form $2^kn$ and $2^km$ where $k,m,n \in \mathbb Z$, with $k > 0$, $m$ and $n$ are both odd, that there are infinitely many numbers in those sequences which differ by 2?
Edit: I am asking whether there are infinitely many terms in those sequences that differ by 2. For instance, if we look infinitely far in those sequences, would we find infinitely many occurrences where $2^km = 2^jn +2$ or $2^km +2 = 2^jn$?
Let $m,n$ be fixed (positive) odd integers, and consider the sequences $\{2^k m\}_{k\ge 1}$ and $\{2^j n\}_{j\ge 1}$.
If $2^{k} m$ and $2^{j} n$ differ by two, then $2^{k-1} m$ and $2^{j-1} n$ will differ by one. Thus one of these is odd and the other is even. Only if $k = 1$ or $j = 1$ is the corresponding term odd, so this cannot happen infinitely many times.
It can happen at most twice, e.g. when $m=1$ and $n=3$ we get two pairs that differ by two:
$$ (4,6) \text{ and } (8,6) $$
The fact that all (even) positive integers appear exactly once in some sequence of the kind originally specified by the OP is occasionally useful. See this recent Problem in Code Golf SE, "The strange ordering of Sharkovskii".