This question amounts to 10na+a=p2, where n, a and are positive integers.
So (10n+1)a=p2. This would require 10n+1 (as well as a) to be a perfect square, and this can never happen, because 11, 101, 1001, 10001 cannot be perfect squares.
If we have a number 10q+1 then square it we get 102q+2×10q+1, which means there's a 2 between the 1s at each end. For example, 112=121, 1012=10201, etc. That's missing in 10n+1, so, no, there are no integer repeats that are perfect squares.