Properties of $a$ and $n$ when $a^n-1$ is prime

Question

In my intro to LaTex course at my university, I've been tasked with investigating the properties of two natural numbers $a$ and $n$ with regards to the formula $m = a^n - 1$ , where $m$ is prime. I am to guess a theorem and then prove it, and write down my work in LaTex as an exercise.

So far I've been pairing the first 1000 primes with $a=1,2,3,4,5,6,7,8$ and $n=2,3,4,5,6,7,8$ and found that the only primes that can be expressed like this are the Mersenne Primes $(p=2^n - 1)$.

I'm therefore thinking that the theorem be "The expression $m = a^n - 1$, where $a, n \in \mathbb{N}$ and $m \in \mathbb{P}$, holds true only when $a = 2$ and $n \in \mathbb{P}$".

However I am looking for help figuring out how to prove this, would you do it by induction or is there maybe a simpler way? Or am I overcomplicating this more than needed and there are straightforward properties of $a$ and $n$ which I'm not seeing. My professor encouraged us to take on any outside help we might get and the focus of the course is writing proofs and learning LaTex. I'm in my first year in what would be my country's equivalent of undergraduate math studies, and I'm currently taking Real Analysis etc, so writing proofs is new but also something I'm doing alot at the moment.

All insight and help appreciated!

Edit: We are to consider the cases of $n > 1$

For clarity I'll put the problem as stated by my professor below (my translation):

"Let $m$ denote a number that can be expressed in the form $m = a^n − 1$, where $a$ and $n$ are natural numbers, and $n > 1$. What can be said about $a$ and $n$ if $m$ is a prime number? Formulate the result of your investigations as a theorem and present the proof of this."

$a^n$ isn't two numbers. $a,n$ may be two numbers, but $a^n$ is one number. — Thomas Andrews, Sep 11 '23 at 13:00
@thomasandrews I've updated the post now, we were to only consider cases where n >1, thanks for pointing it out! — R.F.P, Sep 11 '23 at 13:03
Hint: Think about the geometric series $1 + a + a^2 + \cdots + a^{n-1} = (a^n-1)/(a-1)$. — Ethan Bolker, Sep 11 '23 at 13:06
Look for patterns. Fix $a$, consider lots of $n$ and look for a common factor. For example, is there a number which divides $6-1,6^2-1,6^3-1,6^4-1$? Can you prove that this number divides $6^n-1$ for all $n$? Can you extend this proof? — Mastrem, Sep 11 '23 at 13:06
@R.F.P You didn't edit the post to mention $n>1,$ and, in fact, mention $n=1,2,\dots,8.$ (Oh, I see the dit, after the fact. Change the question, don't add an edit nore.) — Thomas Andrews, Sep 11 '23 at 13:40
Does this answer your question? Show that if $n>1$ and $a$ is a positive integer with $a^n-1$ prime, then $a=2$ and $n$ is prime — Anne Bauval, Sep 11 '23 at 13:41
Such primes are necessarily Mersenne primes. $51$ of them are known , it is conjectured that there are infinite many. — Peter, Sep 11 '23 at 15:40

Properties of $a$ and $n$ when $a^n-1$ is prime

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