Is there a power of 2 (greater than 2^3) that is a palindrome?

Question

Simple question, not sure how to neatly find the answer.

Answer 1

This isn't a solution, but it's easy to see that if such a palindrome exists, it must have an odd number of digits since any palindrome with an even number of digits is divisible by 11. Ie.
10^k+10^(k+2m+1) =
10^k(10^(2m+1)-1) =
10^k(10+1) (10^2m - 10^(2m-1) + 10^(2m-2) - 10^(2m-3) + ... - 1)