I saw this problem recently and it is deceptively hard. The usual $\pmod{4}$ trick won't work, and indeed there will be perfect squares whose last $n$ digits will be $\underbrace{88\cdots8}_{n\text{ times}}1$ , for any $n$.
I can show that if $\underbrace{88\cdots8}_{n\text{ times}}1$ is a perfect square, $2n$ digits total, then the decimal representation of $\sqrt{8/9}$ must start with $n$ digits followed by almost that many zeroes (like $n-1$ or so). This seems preposterous, but I can't figure out how to rule it out. I've thought of using continued fractions, since there is a limit on how well $\sqrt{8/9}$ can be approximated with rationals, but I can't seem to get a tight enough inequality. Maybe someone with more expertise in Diophantine approximation can answer this?
Note: even if you can prove that there aren't infinitely many perfect squares of this form, even if you don't have a computable bound on how large they might be, I'd still be interested.
The number $88\cdots81$ in general form is
$$ (10^n - 1){8 \over 9} - 7 \text{ where } n \text{ is the total number of digits.} $$
If we need this to be a perfect square, we need to solve the Diophantine equation
$$ (10^n - 1){8 \over 9} - 7 = y^2, y \in \mathbb{Z} $$
This implies
$$ y = \pm \bigg({1 \over 3} \sqrt{2^{n + 3} 5^{n} - 71}\bigg) $$
The term under the radical needs to be a perfect square for $y$ to be an integer and also a multiple of $9$. $\text{i.e.,} \exists x \in \mathbb{Z}$ such that,
$$ 2^{n + 3} 5^{n} - 71 = (3x)^2 $$
Case 1 ($n$ is even):
Since $n$ is even, we may take $n = 2m$ and rewrite the equation as
$$ 2^3(10)^{n} - 71 = (3x)^2 \\ \implies 2^3(10)^{2m} - 71 = (3x)^2 \\ \implies 8(10^m)^{2} - 71 = (3x)^2 $$
Taking $10^m = Y, 3x = X$,
$$ 8Y^{2} - X^2 = 71 $$
This is a Pell equation.
Case 2 ($n$ is odd):
Since $n$ is odd, we may take $n = 2m+1$ and rewrite the equation as
$$ 2^3(10)^{n} - 71 = (3x)^2 \\ \implies 2^3(10)^{2m+1} - 71 = (3x)^2 \\ \implies 80(10)^{2m} - 71 = (3x)^2 \\ \implies 80(10^m)^{2} - 71 = (3x)^2 $$
Taking $10^m = Y, 3x = X$,
$$ 80Y^{2} - X^2 = 71 $$
This is also a Pell equation.
We may solve those Pell equations to find solutions for $n$.