Is there a perfect square (other than 9) all of whose digits are 7, 8, or 9?

A short proof for the fact that $\ldots88889$ will appear as the last decimal digits of a square. Consider the modular inverse of $3$ modulo $10^m$. That is, let $n\equiv\ldots 66667$. Then $(3n)^2\equiv1\pmod {10^m}$ and therefore $n^2$ is the modular inverse of $9$. Modulo $10^m$ we have $-1/9\equiv\ldots11111$, so we also have $$n^2\equiv(1/3)^2=1/9\equiv-(\ldots11111)\equiv\ldots88889\pmod{10^m}.$$

Of course, this does not settle the main question, only proving the futility of trying to prove the non-existence of such squares by studying any finite segment of least significant digits.