Does every positive power of $7$ contain $2$ in their ternary representation? I.e. Determine whether there exists a finite subset of non-negative integers $\{a_1 < \dots < a_n \}$ and a positive integer $b$ such that $$ \sum_{i = 1}^n 3^{a_i} = 7^b. $$
I've done a little bit of computation and found that if there are no solutions for $a_i$ if $b < 5000$. So I feel like there is no solution in general and I've had some unsuccessful attempts at proving it. I began by looking at the last few digits of the termary representation of $7^n$, effectively looking at the $7^n \ \textrm{mod } 3^k $ for certain values of $k$. With this I can show is that $7^{3n}$ always contains a $2$. But I realised that for any fixed $k$ there exists $n$ such that $7^n \equiv 1 \ \textrm{mod } 3^k$ since $7$ is a unit in $\mathbb{Z}/ 3^k\mathbb{Z}$. I've also had a look through the computations at the first first few digits of the ternary representation of $7^n$ but I haven't seen any repeating patterms. As a follow up question: Does there exists $k$ such that for all $n$ there is always a $2$ within the first $k$ significant digits of $7^n$?
If anyone knows any results related to this question, I would also be interested to hear about those.