If $n,k\in\mathbb N$, solve $3^k-1=x^n$. If $n,k\in\mathbb N$, solve $$3^k-1=x^n$$
This seems like an interesting problem. I've currently tried a few things (one could try a lot of things in this case):
$1)$ $x^n\equiv -1\pmod 3\Rightarrow x\equiv -1\pmod 3\Rightarrow n=2r+1$, where $r\in\mathbb N\cup\{ 0\}$.
Which brings us to $3^k-1=x^{2r}x$.
$2)$ $2\mid 3^k-1$, so $x^{2r}x$ is divisible by $2$.
$3)$ By checking congruence modulo $4$, we see that $k=2m$, where $m\in\mathbb N$.
Thus $(3^m+1)(3^m-1)=x^{2r}x$. That's just a few things we could try.
 A: You might want to use Zsigmondy's theorem (don't attempt to pronounce this name). This is, in some sense, weaker then Catalan's conjecture so it could be nicer to use this theorem.
An elementary proof can be found here, the required result is Theorem 3. As is written on the first page: ''Except for such basic facts, the paper is self-contained.''
I hope this is somehow satisfactory.
A: Since $n$ is odd, it has some odd prime divisor $p$, or is $1$, if it is $1$, we have an infinite family of trivial solutions. Since if we care of only prime $n$ we are done (why?), let's suppose $n=p$.
$$3^k=(x+1)(1-x+x^2-x^3\cdots)$$
And since $x\equiv 2 \pmod{3}$, the residues on the right parentheses $\pmod3$ look like
$$1+\underbrace{-2+1+-2+1\cdots}_\text{p-1 times}\equiv 1+\underbrace{1+1+1\cdots}_\text{p-1 times}\equiv p \pmod{3}$$
Therefore we want that $p$ is a multiple of $3$, therefore $p=3$
So finally we have $x^2-x+1=3(3q^2-3q+1)$. Since the parentheses is not divisible by $3$, and it must be a power of $3$, it must be $1$. So we have $3q(q-1)=0 \iff q=0,1$. $q=0$ does not work.
Therfore the only solution is $x=3q-1=2$, $n=3$, $k=2$
A: The only solution is $k=2,n=3,x=2$, if $n>1$. We can easily see that this is a consequence of a well known theorem Catalan's conjecture.
