This question was asked at MSE but recieved no attention at all. Here it is:

Are there finitely many $(n,k) \in \mathbb{N}^2$ with $2^n-1=p_1p_2\cdots p_k$ ?

$p_1=3,p_2=5 , ...,p_k$ are consecutive odd primes in ascending order.
An example is when $n=4, k=2$:
$2^4-1=3\cdot 5=p_1p_2$
Are there finitely many $n$?
I tried to use Zsigmondy's theorem without success.
Thanks in advance!

This question was asked at MSE but recieved no attention at all. Here it is:

Are there finitely many $(n,k) \in \mathbb{N}^2$ with $2^n-1=p_1p_2\cdots p_k$ ?

$p_1=3,p_2=5 , ...,p_k$ are consecutive odd primes in ascending order.
An example is when $n=4, k=2$:
$2^4-1=3\cdot 5=p_1p_2$
Are there finitely many $n$?
I tried to use Zsigmondy's theorem without success.
Thanks in advance!