Positive divisors of $P(x,n)=1+x+x^2+ \cdots + x^n$ that are congruent to $1$ modulo $x$ This is a follow-up question to Positive integer solutions to the diophantine equation $(xz+1)(yz+1)=z^4+z^3 +z^2 +z+1$
Let \begin{equation}
 P(x,n)= 1+x+x^2+ \cdots + x^n,  \end{equation}
\begin{equation}
h(m,n)= \frac{m^n-1}{m+1} \end{equation} if $n$ is even and \begin{equation} h(m,n)=\frac{m^{n}-m}{m^2-1}
\end{equation} if $n$ is odd,
where $x, n > 0, m>1$ are non negative integers.
Then;
Claim 1: For all $n>0, m>1$, $m\cdot h(m,n)+1$ divides $P(h(m,n), n)$
Claim 2: Furthermore, for a fixed pair $(m,n),  h(m,n)$ is the largest integer $x$ such that $mx+1 \ | \  P(x,n)$ i.e $mx+1 \ \nmid  \  P(x,n)$ for all $x>h(m,n)$.
These claims are verified for all $n < 10^4, m < 10^2$, a high level of certainty! To prove these claims, my idea would be to use proof by induction but it appears tricky to proceed with induction. How does one go about proving these results? (An elementary proof is preferred or a counterexample).
Claim 2 is particularly interesting because, if its true, then, by just looking at the sizes of $m, n$, one will be able to rule out divisors of $P(x,n)$ of the form $mx+1$.
There’s nothing particularly unique about polynomial $P(x, n)$ and $mx+1$ to limit Claim 2 to $P(x, n) $ or $mx+1$; Only that $P(x,n)$ has no factor $mx+1$ with $m>1$ in its factorization which is why $m$ is set greater than $1$. If $m=1$ is allowed, claim 2 is false as $x+1$ divides $P(x,n)$ for all $x$ when $n+1$ is composite. If Claim 2 is true, the following conjecture is reasonable enough;
Conjecture 1
Let $f(x)$ and $g(x)$ be two polynomials of a non negative integer $x$ with integer coefficients, g(x) does not factor into $ (a/b)h(x)f(x) $ for some integers $a, b \not = 0 $ and polynomial $h(x) $ with integer coefficients. Then, $f(x)$ divides $g(x)$ for finitely many $x$.
 A: Conjecture 1 follows from a (multiple of) Bezout identity for polynomials:
$$mu(x)f(x)+mv(x)g(x)=md(x),$$
where $d(x) :=\gcd(f(x),g(x))$ and a positive integer $m$ is chosen such that polynomials $mu(x)$, $mv(x)$, $md(x)$ have integer coefficients. Then $f(x)\mid g(x)$ for some $x$ implies that $\frac{f(x)}{d(x)}\mid m$, which has a finite number of solutions as soon as $d(x)$ has degree smaller than $f(x)$.
In fact, I used this identity to design an algorithm for finding all $x$ with $f(x)|g(x)$.
Also, when $d(x)\equiv 1$ (i.e., $f(x)$ and $g(x)$ are coprime), the smallest possible $m$ is called the reduced resultant, which can be computed as explained in https://mathoverflow.net/q/333279

ADDED. As for the claims, they follow from the formula for the absolute value of resultant of $P(x,n)$ and $mx+1$ (see A062160) given by
$$P(m,n)=\frac{m^{n+1} - (-1)^{n+1}}{m+1}.$$
It follows that if $(mx+1)\mid P(x,n)$, then $(mx+1)\mid P(m,n)$.
And as soon as $mx+1 > P(m,n)$, i.e., $x > \frac{m^n - 1}{m+1}$, no divisibility can happen.
