How would I go about proving the following:

For any odd positive integer $s$, there exists a sequence of nonnegative integers $( a_0, a_1, \cdots, a_{n-1})$ and a nonnegative integer $m$ such that,

$$ 3^ns + \sum_{k=0}^{n-1} 3^{n-k-1}2^{a_k}=2^m.$$

I am really stuck. I was thinking of using the ring $\mathbb{Z}_{2^m}$ in a proof by contradiction, but I cannot even get started reducing the LHS to something simpler.

Note 1: I have reason to believe there exists such a sequence where

1. $a_0=0$
2. $\{a_n\}$ is strictly monotonically increasing

Note 2: I think an example might help.

$$ n=3 \quad \{a_n\} = [0, 2g-1, 2g+3] \quad s = \frac{5\times 4^g -2}{6} \quad m= 2g+5 \quad g >0$$

Note 3: I originally asked this on the Mathematics Stack Exchange, but it seems to be a question more suited for this exchange.