[Edited to give direct connection with Caporaso-Harris-Mazur via Kevin Buzzard's idea]
Such results are probably true but out of reach of present-day techniques. For each $p$ or $k$, and every sufficiently large $n$ (say $n>k$), the $n$-term arithmetic progressions of the desired type are parametrized by nontrivial points on some algebraic variety, call it $V_n$, of fixed dimension: dimension $2$ if $p$ is fixed (assuming it's not of the form $a(x-x_0)^k+b$ in which case we're back to Darmon-Merel), and degree about $k$ if $p$ is allowed to vary over all polynomials of degree $k$. In each case we have for each $n$ two maps $V_n \rightarrow V_{n-1}$ of degree $k$ that forget the first or last term of the progression; and $V_n$ should be of general type for $n$ large enough. We're now in a setting similar to that of this recent Mathoverflow question (#73346), and I give much the same answer as I did for that question: the claim should follow from the Bombieri-Lang conjectures plus some possibly nontrivial extra work, as in
L.Caporaso, J.Harris, and B.Mazur: Uniformity of rational points, J. Amer. Math. Soc. 10 #1 (1997), 1-45
but (excluding some very special cases that don't seem relevant here) we have no techniques for proving such results unconditionally on varieties of dimension greater than 1.
EDIT Indeed this is a special case of Caporaso-Harris-Mazur, by adapting Kevin Buzzard's observation in his comment on the original question: write the equations as $f(x_m)=m$ $(m=1,2,\ldots,n)$ for some degree-$k$ polynomial $f$ (obtained from $p$ by suitable translation and scaling), and consider just those $m$ in that range which are of the form (say) $y^5$ or $y^5+1$. By Mason's theorem (polynomial ABC) either $f$ or $f-1$ has at least two zeros whose order is not a multiple of $5$, so either $f(x) = y^5$ or $f(x)=y^5+1$ defines a curve of genus at least $2$. But the genus is clearly $O(k)$ for any polynomial $f$ of degree $k$. So, if the number of rational points on such a curve has a uniform bound, then so does the length of an arithmetic progression of values of a polynomial of bounded degree.
Come to think of it, the reduction to C-H-M could also be obtained more directly from the curve $p(x')-p(x)=d$ (for $k>3$), or $p(x''')-p(x'')=p(x'')-p(x')=p(x')-p(x)=d$ (to cover $k=2$ and $k=3$ as well), where $d$ is some multiple of the common difference of the arithmetic progression.
In the case $s=p/q$ with $|p| \geq 3$, there is no $3$-term AP in $P_s$. The proof is by reducing to the case $s=p$, as follows.
Let $A=a^p$, $B=b^p$, $C=c^p$ such that $A^{1/q}+C^{1/q}=2B^{1/q} \quad (*) \quad$ and $(A,B,C)=1$. Let $K=\mathbf{Q}(\zeta_q)$ be the $q$-th cyclotomic field and $L=K(A^{1/q},B^{1/q},C^{1/q})$. Then $L/K$ is a finite abelian extension of exponent dividing $q$ and by Kummer theory, such extensions are in natural bijection with the finite subgroups of $K^{\times}/(K^{\times})^q$. The extension $L/K$ corresponds to the subgroup generated by the classes $\overline{A},\overline{B}, \overline{C}$ of $A,B,C$ in $K^{\times}/(K^{\times})^q$. In view of the following lemma, it suffices to prove $L=K$.
Lemma 1. If $n^p$ is a $q$-th power in $K$, then $n$ is a $q$-th power in $\mathbf{Z}$.
Proof. Assume $n^p=\alpha^q$ with $\alpha \in K$, then taking the norm we get $n^{p(q-1)}=N_{K/\mathbf{Q}}(\alpha)^q$. Since $\alpha$ is an algebraic integer, we get that $n^{p(q-1)}$ is a $q$-th power in $\mathbf{Z}$, and since $p(q-1)$ and $q$ are coprime, we get the result.
Lemma 2. The integer $B$ is relatively prime to $A$ and to $C$.
Proof. By symmetry, it suffices to prove $(A,B)=1$. Let $\ell$ be a prime number dividing $A$ and $B$. Then $\ell^{1/q}$ divides $A^{1/q}$ and $B^{1/q}$ in the ring $\overline{\mathbf{Z}}$ of all algebraic integers. By $(*)$ it follows that $\ell^{1/q} | C^{1/q}$. Thus $C/\ell \in \mathbf{Q} \cap \overline{\mathbf{Z}} = \mathbf{Z}$ which contradicts $(A,B,C)=1$. This proves Lemma 2.
By equation $(*)$, we have $K(B^{1/q}) \subset K(A^{1/q},C^{1/q})$ which reads $\overline{B} \in \langle \overline{A},\overline{C} \rangle$ in $K^{\times}/(K^{\times})^q$. We can thus write $B \equiv A^{\alpha} C^{\gamma} \pmod{(K^{\times})^q}$ for some $\alpha,\gamma \geq 0$. By a reasoning similar to Lemma 1, we deduce that $B/(A^{\alpha} C^{\gamma})$ is a $q$-th power in $\mathbf{Q}$ but since this fraction is in lowest terms (Lemma 2), we get that $B$ is a $q$-th power in $\mathbf{Z}$.
Now let $\sigma$ be an aribtrary element in $\mathrm{Gal}(L/K)$. We have $\sigma(A^{1/q}) = \zeta \cdot A^{1/q}$ and $\sigma(C^{1/q})=\zeta' \cdot C^{1/q}$ for some $q$-th roots of unity $\zeta$ and $\zeta'$. Considering the real parts of both sides of $\sigma(*)$, we see that necessarily $\zeta=\zeta'=1$. This shows that $L=K$ as requested.
Best Answer
Probably not, but a proof is hopeless. Ruzsa and Gyarmati have a preprint in which they construct such a subset of size something like $N/\log \log N$.
Even the colouring version (that is, finite colour the squares, does one of the classes contain a 3-term progression) is open. A very closely-related question (Schur's theorem in the squares) is explicitly asked as Question 11 in this paper by Bergelson:
http://www.math.iupui.edu/~mmisiure/open/VB1.pdf
It is possible to show that a positive density subset of the squares contains a solution to $\frac{1}{4}(x_1 + x_2 + x_3 + x_4) = x_5$ by adapting the technique of arXiv:math/0302311. I'd have to admit this is slightly more than a back of an envelope calculation :-)