$x^m-1 \nmid f(x)$ in $\mathbb{Z}/p\mathbb{Z}[x]$ where $f(x)=(x+1)((x+1)^{2m}+(x+1)^{m}+1)$ Problem: Suppose that $p$ is a prime. Suppose that there is $m \in \mathbb{N}$ such that $p=1+3m$. Define:
$$f(x)=(x+1)((x+1)^{2m}+(x+1)^{m}+1) \in \mathbb{Z}/p\mathbb{Z}[x]$$
I would like to prove that if $p \neq 7$ and $p \neq 13$ then $x^m-1 \nmid f(x)$ in $\mathbb{Z}/p\mathbb{Z}[x]$.
Attempt: I observed that $f(x-1)(x^m-1)=x^p-x$. Then I observed that $x^p-x$ has all its roots in $\mathbb{Z}/p\mathbb{Z}$ and that its roots are all distinct. I would like to prove that for $m \geq 6$ then $x^m-1$ and $((x-1)^m-1)$ share at least a root. This is sufficient because then we would have that $x^p-x$ has at least one multiple roots (I mean that it is not simple).
 A: An element  $a\in{\bf Z}/p{\bf Z}$ is a root of $x^m-1$ if and only is it is a non-zero cube. If it is also a root of $(x-1)^m-1$, then $a+1$ is also a non-zero cube. This leads to a ${\bf Z}/p{\bf Z}$-rational point $P$ on the  projective genus $1$ curve $x^3=y^3+z^3$
with the property that $x,y,z\not=0$. Since there are nine points with $x,y,z=0$, Hasse's theorem implies that $P$ exists whenever $p+1-2\sqrt{p}>9$. In other words, when $p>16$.
A: This is an attempt, please point out if there is an error.
Not a complete answer. Too long for a comment
Consider the polynomial $f(x)=(x+1)((x+1)^{2m}+(x+1)^m+1)$
We get two useful identities, inspired by the proceedings of the OP
$$f(x)((x+1)^m-1)=(x+1)^p-(x+1)\tag{1}$$$$f(x-1)(x^m-1)=x^p-x \tag{2}$$
Let, if possible, $x^m-1\mid f(x)$. Then write $f(x)=(x^m-1)g(x)$ for some $g(x)\in\Bbb{F}_p[x]$. Then $$(x^m-1)((x+1)^m-1)g(x)=(x+1)^p-(x+1)\tag{3}$$ $$(x^m-1)((x-1)^m-1)g(x-1)=x^p-x\tag{4}$$
Therefore, to reach a contradiction, as the OP predicted, it's enough to prove that at least any two among the three $((x+1)^m-1),(x^m-1),((x-1)^m-1)$ share a common root (because for any indeterminate T, the polynomial $T^p-T$ in $\Bbb{F}_p[T]$ has exactly $p$ distinct roots). Well, the elements of $\Bbb{F}_p$ which are not a root of any of the three mentioned, are elements of order three in the multiplicative group $\Bbb{F}_p^{\times}$ of order three. If all these roots of them were distinct, then we would have at most $3m+1-3m=1$ element of order $3$. Indeed, let $$x_1,x_2,\ldots,x_m$$ be the roots of $x^m-1$. Then $$x_1+1,x_2+1,\ldots,x_m+1\\x_1-1,x_2-1,\ldots,x_m-1$$ are the roots of of the rest of the two polynomials $(x+1)^m-1$
This argument works in two cases
Case 1 when $(x+1)^m-1$ and $x^m-1$ share a root
Case 2 When $x^m-1$ and $(x-1)^m-1$ share a common root
Case 3 What to do when $(x-1)^m-1$ and $(x+1)^m-1$ share a common root?
and $(x-1)^m-1$ and all these $3m$ elements are in $\Bbb{F}_p$. But this is a contradiction to Sylow's theorem which says that there exists a subgroup of order $3$ and hence at least $2$ elements of order $3$.
