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 a repunit in base 10 cannot even be a perfect power (so in particular not a cube).

Another relevant reference is by the same authors "On integers with identical digits" containing among other things the result (Thm 2), which in particular gives all repunits that are perfect powers in all (nontrivial) bases up to $10$:

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.

Added: The more general questions to classify all repunits (in arbitrary bases) that are cubes or more generally perfect powers is AFAIK open. It is the question for solutions of the Diophantine equation (with $q=3$ for cubes) $$\frac{x^n - 1 }{x-1} = y^q$$ mentioned in the title of the paper mentioned above, called Nagell--Ljunggren equation.

It is conjecture that the set of all non-trivial (i.e., $n,q \gt 1$) solutions $(x,y,n,q)$ is given by $(3,11,5,2)$, $(7,20,4,2)$, and $(18,7,3,3)$, the former two correpond to the two repunits (that are squares) mentioned in the result I recall above, and the third to the repunit in base $18$ that is a cube mentioned by Gerhard Paseman in a comment.

Yet, as said this is open. But there are numerous partial results, for example:

1. no other than the mentioned ones (nontrivial) repunit is a square

2. if there were another cube then it has at least $29$ digits, and the number of digits is a prime and $5$ mod $6$.

In fact in a different terminology, A geometric series equalling a power of an integer, the Nagell--Ljunggren came up a while ago on MO; there some additional informaition and pointers to literature can be found.

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 a repunit in base 10 cannot even be a perfect power (so in particular not a cube).

Another relevant reference is by the same authors "On integers with identical digits" containing among other things the result (Thm 2), which in particular gives all repunits that are perfect powers in all (nontrivial) bases up to $10$:

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.

Added: The more general questions to classify all repunits (in arbitrary bases) that are cubes or more generally perfect powers is AFAIK open. It is the question for solutions of the Diophantine equation (with $q=3$ for cubes) $$\frac{x^n - 1 }{x-1} = y^q$$ mentioned in the title of the paper mentioned above, called Nagell--Ljunggren equation.

It is conjecture that the set of all non-trivial (i.e., $n,q \gt 1$) solutions $(x,y,n,q)$ is given by $(3,11,5,2)$, $(7,20,4,2)$, and $(18,7,3,3)$, the former two correpond to the two repunits (that are squares) mentioned in the result I recall above, and the third to the repunit in base $18$ that is a cube mentioned by Gerhard Paseman in a comment.

Yet, as said this is open. But there are numerous partial results, for example:

1. no other than the mentioned ones (nontrivial) repunit is a square

2. if there were another cube then it has at least $29$ digits, and the number of digits is a prime and $5$ mod $6$.

In fact in a different terminology, A geometric series equalling a power of an integer, the Nagell--Ljunggren came up a while ago on MO; there some additional informaition and pointers to literature can be found.

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.

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 a repunit in base 10 cannot even be a perfect power (so in particular not a cube).

Another relevant reference is by the same authors "On integers with identical digits" containing among other things the result (Thm 2), which in particular gives all repunits that are perfect powers in all (nontrivial) bases up to $10$:

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.

Added: The more general questions to classify all repunits (in arbitrary bases) that are cubes or more generally perfect powers is AFAIK open. It is the question for solutions of the Diophantine equation (with $q=3$ for cubes) $$\frac{x^n - 1 }{x-1} = y^q$$ mentioned in the title of the paper mentioned above, called Nagell--Ljunggren equation.

It is conjecture that the set of all non-trivial (i.e., $n,q \gt 1$) solutions $(x,y,n,q)$ is given by $(3,11,5,2)$, $(7,20,4,2)$, and $(18,7,3,3)$, the former two correpond to the two repunits (that are squares) mentioned in the result I recall above, and the third to the repunit in base $18$ that is a cube mentioned by Gerhard Paseman in a comment.

Yet, as said this is open. But there are numerous partial results, for example:

1. no other than the mentioned ones (nontrivial) repunit is a square

2. if there were another cube then it has at least $29$ digits, and the number of digits is a prime and $5$ mod $6$.

In fact in a different terminology, A geometric series equalling a power of an integer, the Nagell--Ljunggren came up a while ago on MO; there some additional informaition and pointers to literature can be found.