On the sequence $a(n)=\gcd(2^n-1,\phi(2^n-1))$

Question

For natural $n$, define the sequence

$$ a(n)=\gcd(2^n-1,\phi(2^n-1)) $$

It doesn't appear to be in OEIS and starts $1,1,1,1,9,1,1,1,3,1,9,1,3,1,1,1,27,1,75,49$

Q1 Can we unconditionally prove $a(n)=1$ infinitely often?

(Infinitely many Mersenne primes $M_p$ implies it).

Q2 Can we unconditionally prove $a(n)$ is bounded infinitely often?

Q3 Can we unconditionally prove $a(n)$ is unbounded infinitely often?

Q4 In case $a(n)$ is unbounded infinitely often, is its factorization related to named prime numbers?

Experimental investigation is not very tractable because of expensive factorization.

Answer 1

Q3: certainly yes. Take $n=2\cdot 3^{k}$, then $2^n-1$ is divisible by $3^{k+1}$,thus $3^k$ divides $a(n)$.

As for Q1 and Q2, I do not see how to avoid the scenario when $2^n-1$ is always divisible by a square of a prime.