Is there any palindromic power of $2$?

My question is in the title:

Is it possible to find $n≥4$ such that $2^n$ is a palindromic number (in base $10$)?

A palindromic number is a number which is the same, independently from which side we read it (forwards and backwards), for example $121, 484, 55755$.

My guess is "no". I know that a palindromic number $x$ with even length (i.e. the number of digits $\max\{n : 10^n \mid x\}$ is even) is a multiple of $11$: see here or here or here. In particular, a power of $2$ with an even length can't be a palindromic number.

However, I don't know what to do with the case of an odd length. For instance, if $x=abcdcba$, then $abccba$ is a multiple of $11$, but I don't see how this can help.

Here is a related question. On MathOverflow, related questions are: (1) and (2). Maybe also this thread (since $(1)$ is focused on binary expansion).

I tried with Mathematica and there is no palindromic power of $2$ with exponent $n<10000$:

 palindromeQ[n_] := IntegerDigits[n] === Reverse@IntegerDigits[n];
 For[i = 1, i < 10000, i++, If[PalindromeQ[2^i], Print[i]] ]

Finally, I think that the answer will be the same if we replace $2$ by any integer $n>1$ which is not a multiple of $11$. I don't know how I could prove (even for the case of even length…) that $11^n$ is not a palindromic number for $n≥5$.

Any hint will be helpful. Thank you!

Solution 1:

According to the Wikipedia page on palindromic numbers:

G. J. Simmons conjectured [that] there are no palindromes of form $n^k$ for $k > 4$ (and $n > 1$).

The cited reference is:

Murray S. Klamkin (1990), Problems in applied mathematics: selections from SIAM review, p. 577