Is it true that the equation $10^{n}9m^{3}=1$ has only one positive integer solution, namely $n=m=1$? I can't find the answer. This has an equivalent description that the repunits $R_n = 11\dots1$ are not cubic numbers.

Rather to my surprise I find an entirely elementary proof that $R_1=1$ is the only (decimal) repunit cube, using nothing beyond quadratic reciprocity (namely the formula for the Legendre symbol $(5/p)$). Let us first dispose of the case that $n$ is even, say $n=2k$. This is routine: write the equation $10^n  9m^3 = 1$ as $$ 9m^3 = 10^n  1 = 10^{2k}  1 = (10^k1) (10^k+1). $$ The two factors are relatively prime, and $10^k1$ is a multiple of $9$. Thus once $k>0$ it follows that $10^k1 = 9m_1^3$ and $10^k+1 = m_2^3$ for some $m_1^{\phantom.},m_2^{\phantom.}$ with $m = m_1^{\phantom.} m_2^{\phantom.}$. But then $m_2^3 \equiv 2 \bmod 9$, which is impossible. (The solution $(m,n) = (0,0)$ escapes because in that one case the factor $10^k1$ is zero so there's no condition on $10^k+1$. We could also have used descent, since $10^k1 = 9m_1^3$ would be a smaller solution of the same Diophantine equation; the solution $(m,n) = (0,0)$ escapes this argument because $k=0=n$ so the new solution is no smaller.) The hard case is $n$ odd. For $n=1$ we obtain the known solution $(m,n)=(1,1)$; and there is no solution with $n=3$ because $R_3 = 111$ is not a cube. We may thus suppose $n \geq 5$, and write $n = 2k+3$ with $k$ a positive integer. Now we take the strange step [see comment at bottom] of adding $9$ to both sides of the equation $9m^3 = 10^n  1$, and writing the result as $$ 9(m+1)(m^2m+1) = 9m^3 + 9 = 10^n + 8 = 10^{2k+3} + 8 = 8 \left(125(10^k)^2 + 1 \right). $$ Thus if $p$ is any odd prime factor of $m+1$ then $p \neq 5$ and $25 \cdot 10^k$ is a square root of $5 \bmod p$, so $(5/p) = +1$ and by quadratic reciprocity $p$ is one of $1,3,7,9 \bmod 20$. As usual, this set of residues $\bmod 20$ is closed under multiplication, so we conclude that any odd factor of $m+1$, prime or not, is one of $1,3,7,9 \bmod 20$. In particular this is true of $(m+1)/2^f$, where $2^f$ is the largest power of $2$ dividing $m+1$. But since $n > 3$ (this is how the solution $(m,n)=(1,1)$ escapes the coming contradiction) we have $f=3$ because $125(10^k)^2+1$ is odd (as is the complementary factor $9(m^2m+1)$ on the lefthand side). Moreover, once $n \geq 5$ we have $R_n \equiv 11111 \equiv 7 \bmod 2^5$, so the putative cube root $m$ of $R_n$ would be $23 \bmod 2^5$, making $(m+1)/8 \equiv 3 \bmod 4$. Since also $m \equiv 1 \bmod 5$ we'd conclude that $(m+1)/8 \equiv 4 \bmod 5$ and thus $(m+1)/8 \equiv 19 \bmod 20$. Since this is not among the four allowed residues we are done. QED The same approach deals with some other cases of the repunitpower problem, but does not settle it completely. For example, in the decimal case $9m^q+1 = 10^n1$ has no solution for any $q \equiv 3 \bmod 4$ (using descent to reduce to the case of odd $n$, and then factoring $9(m^q+1) = 10^n+8$ as before). There are no nontrivial solutions for $2q$ or $5q$ (by reduction mod $2^2$ and $5^2$ respectively), and $q=9$ is a special case of $q=3$, so $q=13$ is the first exponent (other than $q=1$...) that we cannot exclude this way. about the "strange step": this feels very artificial, though I'm not familiar enough with the literature on this NagellLjunggren equation or related problems to tell if it's a standard technique. The only other time I remember such a thing working is for a problem that I concocted for the purpose 30+ years ago: Prove that the Diophantine equation $$ y^2 = 7x^2+8x3 $$ has no positive integer solutions. There are infinitely many solutions with $x<0$, such as $(x,y) = (2,3)$, so there's no easy congruence argument (though the problem is routine using the theory of Pell's equation). However, adding $9x^2$ to both sides yields $(3x)^2 + y^2 = (4x1)(4x+3)$, at which point the twosquare theorem soon produces a contradiction (NB if $x<0$ then $\left4x1\right$ and $\left4x+3\right$ are $+1 \bmod 4$ !). 


As expected $(m,n)=(1,1)$ is the only solution in positive integers of the exponential Diophantine equation $10^n  9m^3 = 1$. An entirely elementary proof of this seems unlikely because any finite list of congruence conditions on $n$ would have to allow $n=1$ (and also $n=0$, which corresponds to the empty repunit $0$), and thus could not exclude infinitely many other potential $n$. A routine but nonelementary solution is to reduce to the three cubic Thue equations $p^39m^3=1$, $10p^39m^3=1$, and $100p^39m^3=1$ (where $p = 10^{\lfloor n/3 \rfloor}$), and then use an effective and practical algorithm for solving such equations. gp takes only a few milliseconds to process each of
and reports that the third equation has no solutions, the second only $(p,m)=(1,1)$, and the first only $(2,1)$ and $(1,0)$. Since neither $2$ nor $1$ is a positive power of $10$ we're done. P.S. I see that quid linked to a page that in turn includes a link to much the same proof except that the third case (which has no solution) is disposed of by elementary considerations, namely reduction $\bmod 13$ (and as it happens there's a similar proof $\bmod 7$). 


It is know by work of Bugeaud and Mignotte "Sur l'équation diophantienne $(x^n  1)/(x1)=y^q$, 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 $10$:
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://wwwirma.ustrasbg.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) $$\frac{x^n  1 }{x1} = y^q$$ mentioned in the title of the paper mentioned above, called NagellLjunggren equation. It is conjecture that the set of all nontrivial (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:
In fact in a different terminology, A geometric series equalling a power of an integer, the NagellLjunggren came up a while ago on MO; there some additional informaition and pointers to literature can be found. 


$\frac {10^n1} 9$ is not a cube: Since a cube is either 0, 1 or 1 mod 9, the number of digits in the repunit is 0, 1 or 1 mod 9. If the number of digits is 1 mod 9, you can show it's not a cube by reducing mod 19. If the number of digits is 1 mod 9, reduce mod 7 to show it's not a cube (except for 1). Suppose $\frac {10^{9n}1} 9$ is a cube. $n =2^k m$ where $m$ is odd. $\frac {10^{9n}1} 9$ factors as $(10^{2^{k1}9m}+1)...(10^{9m}+1)\frac {10^{9m}1} 9$ and the factors are relatively prime, so each must be a cube. The factors that have a power of 10 plus 1 can't be cubes because a cube + 1 isn't a cube. Reducing the last factor mod 19, $10^9\equiv 1 (19)$ so $\frac {10^{9m}1} 9\equiv \frac {2} 9\equiv 4 (19).$ $4^{18/3}=4^6\equiv 11 (19)\neq 1 (19)$ so this factor also is not a cube. So the product can't be a cube. So no repunit base 10 is a cube. 

