This question was asked at [MSE][1] 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][2]
without success.  
Thanks in advance!


  [1]: https://math.stackexchange.com/questions/1465274/is-2n-1-finitely-many-times-the-product-of-consecutive-primes
  [2]: https://en.wikipedia.org/wiki/Zsigmondy%27s_theorem