Constraints on equations $(a^n-1)\equiv 0 \mod (b^n-1)$

Question

When I first read this question (Conjecture about the system of powerful equations) about systems of equations where $(a^n-1)\equiv 0 \mod (b^n-1)$, I assumed the question indicated the modular relationship was to be true for every $n\in \mathbb N$. Reading the comments, however, it became clear that the question was about systems where the modular relationships are true for all $n\le m$.

If $(a^n-1)\equiv 0 \mod (b^n-1)$ for all $n$, $a=b^k$ is one possibility, as that would result in $((b^n)^k-1)\equiv 0 \mod (b^n-1)$ which is true for all $n$. However, for the narrower case where $n\le m$ (in particular $m=2$, so $n=1,2$), an example was given where $(a^n-1)\equiv 0 \mod (b^n-1)$, but $a\ne b^k$ ($a=5,\ b=3$).

This raises two related questions:

1. In the case where $n$ has no upper limit, are there solutions to the system of equations $\forall n: (a^n-1)\equiv 0 \mod (b^n-1)$ other than $a=b^k$?

2. If not, then is there a value of $m$ above which there are no other solutions than $a=b^k$?

Answer 1

Reference for part $1$ see here (the first answer). From my proof (the second answer) we get that it's enough to assume that $m = a$ and it gives some answer for the part $2$. In fact the same proof allows us to formulate the next motivational theorem (this theorem can be seen as a motivation to the Conjecture).

$\bf{Theorem.}$

Let given a system of modular equations

$$\left\{\begin{array}{l}a-1 \equiv 0\mod{b-1}\\a^2-1 \equiv 0\mod{b^2-1}\\ \ldots\\a^n-1 \equiv 0\mod{b^n-1}\end{array}\right. (\star)$$

Then $\frac{(b-a)(b^2-a)\cdot\ldots\cdot(b^n-a)}{(b-1)(b^2-1)\cdot\ldots\cdot(b^n-1)} = \frac{u}{v}$, where $v$ the divisor of $n!$.