If N = 4.5 x 10^m, then there are four such strings of zeroes, none of which can be at the center.
Take the longest string of zeroes. There can be s of them, up to s=9.
They can’t end the string. The first string must palindrome with the last string. The first string must be preceded by a 1, the last string is followed by an s. So, s=1, and there is only one string of m zeroes, as you said.
Now, where does your proof break down for my counterexample? The non-equivalence of 1 and the base minus one?
Looks like you’ve successfuly patched my proof. The proof doesn’t apply to base 2 because in that case
(1) s is always 1, so it reduces to the situation I addressed in my incomplete proof, in which case,
(2) in base 2 it is possible for the last digit of the number 2^n - 1 and the first digit of 2^n to be the reverse of the first two digits of 2^n + 1, whereas my proof depends on this not being possible.
To complete the proof for base two, s still has to be 1, since that is the only choice in that base. Then the string of m zeroes in the middle is preceded by m+1 ones, and followed by a one plus m-1 zeroes. No palidrome is possible then unless m=1. So the only possible palindrome in base two is the one I gave, 11011, counting from one to three.