Vtyjkyuk

I enumerated the number of Misiurewicz points using SageMath to factor into irreducible polynomials over $mathbbZ$, where the degree (after discarding factors corresponding to lower (pre)periods) gives the number of roots over $mathbbC$ (which are guaranteed to have multiplicity $1$). Then I played around trying to fit a function to the data in Haskell, and came up with this $M(q,p)$ that counts the number of Misiurewicz points in the Mandelbrot set:

$$
beginaligned
H :mathbbN^+ &to mathbbN \
H (p) &= sum_d mid p muleft(fracpdright) 2 ^ d - 1 \
M :mathbbN^+ times mathbbN^+ &to mathbbN \
M (q, p) &= left(2 ^ q - 1 - mathbf1_q equiv 1 mod pright) H (p)
endaligned
$$

$mu$ is the Moebius function.
$mathbf1$ is the indicator function. H(p) counts hyperbolic components with preperiod $0$.
$q$ is the desired preperiod, $p$ is the desired period.

However while this fits all the data I have available ($q + p le 16$), and I know that $H$ is correct, I have no proof of correctness of $M$. Any ideas how to go about proving this? In all my tests I ended up with just $1$ irreducible polynomial, with multiplicity $1$, after discarding lower (per)period factors. But tests are not a proof.

Misiurewicz Points for Polynomial Maps and Transversality by Benjamin Hutz, Adam Towsley

Corollary 3.3. The number of $(m,n)$ Misiurewicz points for $f_d,c$ is
$$
M_m,n = begincases
sum_k mid n muleft(n over k right) d^k-1 & m = 0 \
(d^m - d^m-1 - d + 1)
sum_k mid n muleft(n over k right) d^k-1 & m ne 0 text and n mid (m - 1) \
(d^m - d^m-1)
sum_k mid n muleft(n over k right) d^k-1 & textotherwise
endcases
$$

There is a typo in the PDF statement of the corollary ($k mid d$ in place of $k mid n$), but its proof uses $k mid n$, so the above is not a verbatim quote.

They have $f_d,c(z) = z^d + c$, so this result is more general than the case $d = 2$ I asked about.