The Boolean algebra generated by sets of prime divisors of the numbers $2^n-1$ Let $\Pi$ be the set of odd prime numbers and let $\mathcal P(\Pi)$ be the Boolean algebra of subsets of $\Pi$. 
For a number $x$ denote by $\Pi(x)$ the set of odd prime divisors of $x$.

Problem. Does each singleton $\{p\}\subset \Pi$ belong to the Boolean algebra generated by the family $\{\Pi(2^n-1):n\ge 2\}$ in $\mathcal P(\Pi)$?

We can also ask a stronger 

Question. Is it true that for every odd prime number $p$ there exit numbers $n_1,\dots,n_k$ such that $\{p\}=\bigcap_{i=1}^k\Pi(2^{n_i}-1)$?


Added in Edit. The number $p=5$ mentioned in the answer of Ofir Gorodetsky gives a counterexamle to Question but not to Problem as $\{5\}=\Pi(2^4-1)\setminus\Pi(2^2-1)=\{3,5\}\setminus\{3\}$.
 A: No. The gcd (greatest common divisor) of $\{ 2^{n_i}-1\}_{i=1}^{k}$ can be seen to be $2^{\gcd(n_1,\ldots,n_k)}-1$, by an application of the Euclidean algorithm. If $p$ divides this expression, then $\gcd(n_1,\ldots,n_k)$ is divisible by the order of $2$ in the multiplicative group $(\mathbb{Z}/p\mathbb{Z})^{\times}$. In general, $2^{\mathrm{ord}_p(2)}-1$ may have prime factors other than $p$.
For instance, if $p=5$ and $p$ divides all of $\{ 2^{n_i}-1\}_{i=1}^{k}$ then each $n_i$ is divisible by $\mathrm{ord}_p(2) = 4$, and so each $2^{n_i}-1$ is divisible by $2^4-1=15= 3 \times 5$.
In fact, the answer to your question is positive only for primes $p$ such that $p^k$ is of the form $2^q-1$ for some $k$, where $q$ is a prime. (For $k=1$ these are known as Mersenne primes.)
A: For example, if $2^n-1$ is divisible by 43, then $n$ is divisible by 14 and therefore $2^n-1$ is divisible by 127. Thus 43 is a counterexample.
A: This was already answered in comments, so I'm posting cw.
The Boolean subalgebra $A$ defined by the OP being generated by finite subsets, it is atomic. The atoms are precisely the (finite) fibers of the function $f$ mapping the odd prime $p$ to the order $f(p)$ of $2$ modulo $p$; it is given by OEIS:A014664.
And indeed $f$ is not injective, so $A$ does not include all singletons. For instance $f(23)=f(89)=11$ (it was mentioned in a comment by Gehrard Paseman).
Hence the problem (and hence the question) has a negative answer. However the question already has a simpler negative answer, as the singleton $\{5\}$ (which belongs) to the Boolean algebra $A$ is not in the smaller set of subsets defined in the question using only finite intersections, see Ofir's answer.
