I tried with mathematics induction, but I couldn't prove.
The difference between consecutive perfect primes $k^2$ and $(k+1)^2$ is $2k+1$.
The difference between $n10^n$ and $(n+1) 10^n$ is $10^n$. Suppose the highest perfect square before the interval is $n10^n$ itself. Then the difference to the next perfect square will be $\sqrt(n10^n)*2+1 = 2\sqrt(n)*\sqrt(10)^n+1$.
This will be smaller than $10^n$ for all $n>=1$. So there must be at least one perfect square in the said interval.