Thinking and searching about it also the case of cubes seems not clear to me, and thus I "upgrade" my comment.
It is know by work of Bugeaud and Mignotte "Sur l'équation diophantienne $(x^n - 1)/(x-1)=y^q$, II" (see Thm 5) that such a number is not a perfect power (not just not a cube).
Another relevant reference is by the same authors "On integers with identical digits" containing among other things the result (Thm 2):
Let $a$ and $b$ be integers with $2 \le b \le 10$ and $1 \le a \le b-1$. The integer $N$ with all digits equal $a$ in base $b$ is not a perfect power, except for $N=1,4,8,9$, for $N=11111$ written in base $3$, for $N=1111$ written in base $7$, for $N=4444$ written in base $7$.
There are earlier contributions to this problem by others, see the papers for references. (Both papers are freely available online on Bugeaud's webpage http://www-irma.u-strasbg.fr/~bugeaud/publi.html see year 1999)
It seems there was some discussion of this question on the Mersenne forum a while ago http://www.mersenneforum.org/showthread.php?t=16295 for cubes specifically and also a proof was given (scroll down a bit). I did not study it in detail, but if you want just the result for cubes it seems more accessible, yet it also involves solving Thue equations.