Consider all n-digit positive integers.

There are 9*10^(n-1) n-digit positive integers

There are roughly 10^(n/2) - 10^((n-1)/2) n-digit squares

Of these, statistically, 0.9^n are expected to be zeroless.

As we increase from n digits to (n+1) digits, the number of squares increases by a factor of √10 but the probability of zeroless only decreases slightly being multiplied by a factor of 0.9.

So going from n digits, to (n+1) digits, the number of zeroless squares is expected to increase by a factor of 0.9*√10 or about 2.8

Stating this as a proof by induction, in the base case, there are 3 one-digit squares containing no zeros.  Assume there are M n-digit zeroless squares.  The above discussion shows that there are roughly 2.8*M zeroless squares which have n+1 digits.