[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
Starting from the equations in my previous answer, we get, by multiplying them in pairs, $$(x-y)x(x+y)(x+2y) + (x-y)x + (x+y)(x+2y) + 1 = (z_1 z_6)^2\,,$$ $$(x-y)x(x+y)(x+2y) + (x-y)(x+y) + x(x+2y) + 1 = (z_2 z_5)^2\,,$$ $$(x-y)x(x+y)(x+2y) + (x-y)(x+2y) + x(x+y) + 1 = (z_3 z_4)^2\,.$$ Write $u = z_1 z_6$, $v = z_2 z_5$, $w = z_3 z_4$ and take differences to obtain $$3 y^2 = u^2 - v^2 \qquad\text{and}\qquad y^2 = v^2 - w^2\,.$$ The variety $C$ in ${\mathbb P}^3$ described by these two equations is a smooth curve of genus 1 whose Jacobian elliptic curve is 24a1 in the Cremona database; this elliptic curve has rank zero and a torsion group of order 8. This implies that $C$ has exactly 8 rational points; up to signs they are given by $(u:v:w:y) = (1:1:1:0)$ and $(2:1:0:1)$. So $y = 0$ or $w = 0$. In the first case, we do not have an honest AP ($y$ is the difference). In the second case, we get the contradiction $abcd + ad + bc + 1 = 0$ ($a,b,c,d$ are supposed to be positive). So unless I have made a mistake somewhere, this proves that there are no such APs of length 4.
Addition: We can apply this to rational points on the surface. The case $y = 0$ gives a bunch of conics of the form $$x^2 + 1 = z_1^2, \quad z_2 = \pm z_1, \quad \dots, \quad z_6 = \pm z_1\,;$$ the case $w = 0$ leads to $ad = -1$ or $bc = -1$. The second of these gives $ad + 1 < 0$, and the first gives $ac + 1 = (a^2 + 1)/3$, which cannot be a square. This shows that all the rational points are on the conics mentioned above; in particular, (weak) Bombieri-Lang holds for this surface.