Is $2^n -1$ finitely many times the product of consecutive primes? 
Are there finitely many $(n,k) \in \mathbb{N}^2$ with 
  $$2^n-1=p_1p_2\cdots p_k$$ 
  where $p_1=3,p_2=5 ,\dots,p_k$ 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!
 A: Any product of consecutive odd primes (starting with $3$ and containing more than two terms) is ending with $5.$  So the last digit of $1+\prod p_k$ is clearly $6.$
Therefore, if this number is a power of two, then it should be a form of $2^{\color{Red}{4k}}=16^k.$
The digit before the last digit of a power of $16$ is varying as $$\color{Green}{1\to5\to9\to3\to7}\to1\to5\to\cdots$$
For obtain more patterns I have calculated few values of $1+\prod p_k.$
$$3\times5=15$$
$$3\times5\times7=105$$
$$3\times5\times7\times11=1155$$
$$3\times5\times7\times11\times13=15015$$
$$3\times5\times7\times11\times13\times17=255255$$
$$3\times5\times7\times11\times13\times17\times19=4849845$$
$$3\times5\times7\times11\times13\times17\times19\times23=3234846615$$
$$3\times5\times7\times11\times13\times17\times19\times23\times31=100280245065$$
A divisibility rule for $16$ which may be useful.
A: With the restriction that $p_1=3$, yes. By Zsigmondy, the $p_i$'s go at least up to $n-1$, so we have
$$\log (2^n-1)>\log\left(\frac12\prod_{p< n}p\right)$$
yet using the prime number theorem (in particular, the form involving the first chebyshev function), the RHS is $n+o(n)$; while the LHS is $n\log 2+O(1)$.
Without the restriction that $p_1=3$, the question is related to the infinitude of Mersenne primes, so probably becomes hard. What the above tells us is that if $n$ yields a counterexample, the least prime divisor of $2^n-1$ has to be $\geq n(1-\log 2)+o(n)$.
