Are there nontrivial perfect powers of integers that are nontrivial repdigits?

Question

For example, $6^5=7776$ is close, but not quite a repdigit. Heuristically, it seems to me that there should not be any, because the longer a number with (effectively) random digits is, the less the odds of that number being a repdigit become. So as the integer power in question gets bigger, the odds of it being a repdigit fall off quickly.

Answer 1

I can eliminate a few cases. Repdigits look like $a(111\cdots1)$. If $a=2$, then you have

$$2(111\cdots1) = b^k.$$

Since the second factor is odd, at most one factor of $2$ appears in $b^k$, a contradiction since $k>1.$

Similarly $a$ can't be $6.$

If $a=4$, then similar reasoning forces $k=2$ and $b=2c.$ Then you have

$$4(111\cdots1) = (2c)^2$$

$$111\cdots1 = c^2$$

which is impossible the left side is congruent to $3$ modulo $4.$

If $a=8$, it (similarly) forces $k=3$ the equation reduces to

$$111\cdots1 = c^3.$$

This was proved impossible here:

https://mathoverflow.net/questions/132011/can-repunits-be-perfect-cubes

So $a$ can't be even.

Now things get fuzzy. If $a$ is odd and prime, it seems like writing

$$p(111\cdots1) = b^k$$

should get us somewhere. Certainly it shows that $p^{k-1}$ divides $111\cdots1.$ But I'm stuck there. That would eliminate $3, 5,$ and $7$.

That's all I got.