Let $x,n,y,q$ be integers greater than one with $n>2$ as well (this is to avoid trivial solutions). The closed form for the geometric series $S(x,n)=\sum_{k=0}^{n-1} x^k$ is $\frac{x^{n}-1}{x-1}$. I'm wondering for which $x,n$ does this yield a perfect power, $S(x,n)=y^q$.
Known perfect powers: I tested for $x\le 50, n\le 30$ and found $S(3,5)=11^2$, $S(7,4)=20^2$, and $S(18,3)=7^3$, but I couldn't find any others. The Mathematica code I used (not saying this is optimal at all; FactorInteger becomes unwieldy for large integers) is below. I don't know if there is a way to test if a number is a perfect power without rote factorization, besides congruence arguments which seem very ad hoc, though perhaps there is something there. This is probably nothing more than a lack of data, but it is worth noting that of the numbers I tested, for $n=3,4,5$ there was only one value of $r$ that gave a perfect power, and none for $n\ge 6$.
Special cases: the case $x=2$ is interesting because Mihailescu's Theorem tells us that $3^2$ and $2^3$ are the only consecutive perfect powers; in particular, $2^{n}-1$ is never a perfect power, so $x=2$ yields no solutions. I actually found the pair $(7,4)$ by hand, because when $n=4$, $S(x,4)=(x^2+1)(x+1)$, and if both of these factors are two times a square, then $S(x,4)$ will be a perfect square as well. You can choose $x^2+1$ to be twice a square by looking at the convergents of $\sqrt{2}$, but of these that I explored further, $7$ was the only one where it plus one was also double a square. This is ad hoc (there's nothing special about being twice a square: being five times or twenty-one times, or whatnot, a square would also work) but perhaps this line could be continued. $n=3,q=2$ has no solutions because of the binomial expansion, and similarly I think it's not bad to see if $n=q+1$ there aren't solutions. When $n=5,q=2$, we have $x(x+1)(x^2+1)=y^2-1$, or $(x^2+x)(x^2+1)=(y+1)(y-1)$; this implies $x$ and $1$ are two apart, i.e. $x=3$, and then $y=11$, and so this is the only solution.
Mathematica code, for what it's worth
S=(1-x^(n))/(1-x);
power[z_] := GCD @@ (FactorInteger[z][[All, 2]])
Table[ power[S]-1,{x,2,30},{n,3,51}];
sa0 = SparseArray[%];
{1,2}+#&/@sa0["NonzeroPositions"]
S/.{x->#[[1]],n->#[[2]]}&/@%
