In connection with this MO post (and without any applications / motivation whatsoever), here is an apparently difficult - but nice - problem.

For a non-zero real number $s$, consider the infinite sequence $$ P_s := \{ 1^s, 2^s, 3^s, \ldots \} . $$ What is the number of terms, say $l(s)$, of the longest arithmetic progression contained in this sequence?

For instance, since there exist three-term arithmetic progressions in squares, but no such four-term progressions, we have $l(2)=3$.

It is easy to see that if $s$ is the reciprocal of a positive integer, then
$P_s$ contains an infinite arithmetic progression; hence we can write
something like $P(1/q)=\infty$. It can be shown that this is actually *the
only case* where $P_s$ contains an infinite progression. (This is certainly
non-trivial, but not that difficult either - in fact, in a different form
this was once posed as a problem on a Moscow State University math
competition).

If now $s$ is the reciprocal of a *negative* integer, then $P_s$ contains an
arithmetic progression of any preassigned length; this is a simple exercise.
Are there any other values of $s$ for which $l(s)$ is infinite?

Conjecture.For any real $s\ne 0$ which is not the reciprocal of an integer, the quantity $l(s)$ is finite; that is, there exists an integer $L>1$ (depending on $s$) such that $P_s$ does not contain $L$-term arithmetic progressions.

Three-term progressions are not rare; say, for any integer $1<a<b<c$ with $b>\sqrt{ac}$ there exists $s>0$ such that $\{a^s,b^s,c^s\}$ is an arithmetic progression. However, I don't have any single example of a four-term progression in a power sequence (save for the case where the exponent is a reciprocal of an integer).

Is it true that $l(s)\le 3$ for any $s\ne 0$ which is not the reciprocal of an integer?

Indeed, excepting the cases mentioned above and their immediate modifications, I do not know of any $s$ such that $P_s$ contains two distinct three-term progressions.

Is it true that if $s\ne p/q$ with integer $q\ge 1$ and $p\in\{\pm1,\pm2\}$, then $P_s$ contains at most one three-term arithmetic progression?

$$ $$

As a PS: I was once told that using relatively recent (post-Faltings) results in algebraic number theory, one can determine $l(s)$ for $s$ rational. Can anybody with the appropriate background confirm this?