I have not thought about this particular problem yet, but let me prove somewhat related claim that if $x,y$ are integers and $x^n-1$ divides $y^n-1,$ $y>1$ for all $n\in\mathbb{N}$ (this condition can be eventually relaxed) then $y=x^m$ for some $m\in\mathbb{N}.$ This, in particular implies that both $x$ and $y$ cannot be primes simultaneously.
The key is the following useful lemma.
Lemma. If $x_1....x_m\ge 1$ are distinct rational numbers and $\alpha_1,...\alpha_m\ne 0$ are real numbers such that $\lim_{n\to\infty}\sum_{k=1}^m\alpha_kx_k^n-m_n=0$ where $m_n\in\mathbb{N},$ then $x_i\in\mathbb{Z},$ for $1\le i\le m.$
Proof. Induction on $m.$ The base case is fairly interesting exercise, I will leave it for the reader. To prove a step of induction from $m$ to $m+1,$ we fix some $x_l=\frac{p}{q}.$ Observe, that
$$\lim_{n\to\infty}\sum_{k=1}^mp\alpha_kx_k^n-pm_n=0$$
and
$$\lim_{n\to\infty}\sum_{k=1}^mq\alpha_kx_k^{n+1}-qm_{n+1}=0.$$
Subtracting last two identities leads to
$$\lim_{n\to\infty}\sum_{k=1}^m\alpha_kx_k^n(qx_k-p)-(qm_{n+1}-pm_n)=0.$$
Since $x_lq-p=0,$ we can apply induction hypothesis to conclude that $x_i\in\mathbb{N}$ for $i\ne l.$ We are left to note that $l$ was chosen arbitrary and thus, $x_i\in\mathbb{N}$ for $1\le m+1.$
Now, in order to apply this to our problem we pick such $m$ that $x^m\le y<x^{m+1}$ and take $m_i=\frac{y^i-1}{x^i-1},$ $x_i=\frac{y}{x^i},$ $\alpha_i=1$ for $1\le i\le m.$ Then
$$m_n-\sum_{k=1}^m\alpha_kx_k^n=\frac{y^n-x^{mn}}{x^{mn}(x^n-1)}\to 0.$$
So all $x_i$ are integers. Thus, $$\frac{y^n-x^{mn}}{x^{mn}(x^n-1)}=0$$
for sufficiently large $n.$ Therefore $y=x^m.$
Note, that we can request $\lim_{n\to\infty}\sum_{k=1}^m\alpha_kx_k^n-m_n=0$ along some "fairly nice" subsequence $n_k.$ Thus, we do not require divisibility $x^n-1$ and $y^n-1$ for all $n.$
Remark 1. As to the second question, I do not think that there is something special about $n,$ in the sense that it can be fairly large. Good computer search should help to find a counterexample.
Remark 2. Starting from any prime $q$ and any $a>1,$ and $q^n-1$ divides $a^n-1$ we can take $x_m=a+m(q^{n}-1)$ and apply Dirichlet's theorem for arithmetic progressions to find infinitely many primes that satisfy our condition.
I cannot see either the motivation or the importance behind studying special numbers $($such as Fermat primes, Mersenne primes$)$.
$a^n-1$ is always divisible by $a-1$, and hence non-prime, or composite $\ldots$ Oh, wait ! Unless $a-1$ $=1\iff a=2$. $($This explains the mathematical interest in Mersenne primes$)$. Also, $a^n+1$ is always divisible by $a+1$, and hence non-prime, or composite $\ldots$ unless $n=2^k$. $($This explains the mathematical interest in Fermat primes; see also$)$.
Why are we studying this? Is it just a purely number theoretic question?
See my answer to this question.
Best Answer
Using Fermat's little theorem $p$ divides $2^p-2$ and if we add your hypothesis $p|2^p+1$ then $p$ divides $3=(2^p+1)-(2^p-2)$