How many $N$ of the form $2^n$ are there such that no digit is a power of $2$? [duplicate]

How many $N$ of the form $2^n,\text{ with } n \in \mathbb{N}$ are there such that no digit is a power of $2$?

For this one the answer given is the $2^{16}$, but how could we prove that that this is the only possible solution? and what about the general case of $x^n, \text{ with } x,n \in \mathbb{N}$?

Solution 1:

Define the acceptable digits to be 0, 3, 5, 6, 7, and 9; and define the score of a number $n$ to be the number of trailing acceptable digits in the decimal expansion of $n$ (with no leading zeroes). So for instance 65536 has a score of 5, and $2^{96} = 79228162514264337593543950336$ has a score of 7 (and this is the smallest power of $2$ with a score greater than 5).

I did a computer search for high-scoring powers of $2$ up to $2^{332192}$ (i.e. those with less than 100000 decimal digits). The highest-scoring was $2^{57072}$, with a score of only 25 (it ends with ...40535076966633036050333696).

So your conjecture is plausible, but it looks like one of those things that could be very difficult to prove.