Let $f(x) = x^d + c$ be a polynomial where $ c \in \mathbb{Q}, d \in \mathbb{N}, d \geq 2 $. For $ n \in \mathbb{N}$, the $n^{\text{th}}$ iterate of $f$ is defined recursively by
$$ f^n (x) = \left\{ \begin{array}{ll} f(x) & \mbox{if } n = 1 \\ f(f^{n-1} (x)) & \mbox{if } n \geq 2 \end{array} \right. $$
A point $P \in \mathbb{Q}$ is said to be a rational periodic point of period $N$ if $f^N (P) = P$ and $N$ is the smallest value for which this is true.
My question is whether there are any known examples of such polynomials with rational points of period greater than 3?
For the case $d=2$: Walde and Russo provided explicit formulae for producing infinitely many polynomials with rational periodic points of periods 2 and 3; Morton showed that there are no polynomials with rational periodic points of period 4; Flynn, Poonen and Schaeffer showed the same for period 5; Poonen conjectured that there can be no rational periodic points with period $N \geq 4$.
For the general case when $d \geq 2$, Sadek has shown that the density of polynomials with a rational periodic point among all polynomials of form $f(x) = x^d + c, c \in \mathbb{Q}$ , is zero. He also provides some necessary conditions that must be satisfied by $c$ and $d$ in order for a polynomial to possess a rational periodic point of period $N$, but does not provide examples.
Examples of polynomials with rational periodic points of period 2 and 3:
$f(x) = x^2 - \frac{21}{16} ,P=\frac{-5}{4}$
$f^1 \left(\frac{-5}{4}\right) = f \left(\frac{-5}{4}\right) = \frac{25}{16} - \frac{21}{16} = \frac{1}{4} $
$f^2 \left(\frac{-5}{4}\right) = f\left(\frac{1}{4} \right) = \frac{1}{16}- \frac{21}{16} = \frac{-5}{4} $
$f(x) = x^2 - \frac{29}{16} ,P=\frac{-7}{4}$
$f^1 \left(\frac{-7}{4}\right) = f \left(\frac{-7}{4}\right) = \frac{49}{16} - \frac{29}{16} = \frac{5}{4} $
$f^2 \left(\frac{-7}{4}\right) = f\left(\frac{5}{4} \right) = \frac{29}{16}- \frac{29}{16} = \frac{-1}{4} $
$f^3 \left(\frac{-7}{4}\right) = f\left(\frac{-1}{4} \right) = \frac{1}{16}- \frac{29}{16} = \frac{-7}{4} $