$\let\dvds\mid$Yes. By Zsigmondy's theorem, $2^{12}-1$ has some prime divisor $p_s$ not dividing $2^i-1$ for $i<12$ (in fact, $p_s=13$). Now, if $k\geq s$, then $p_s\dvds 2^n-1$, so $12\dvds n$ and hence $3^2\dvds 2^n-1$, which is impossible.

$\let\dvds\mid$Yes. By Zsigmondy's theorem, $2^{12}-1$ has some prime divisor $p_s$ not dividing $2^i-1$ for $i<12$ (in fact, $p_s=13$). Now, if $2^n-1=p_1p_2\dots p_k$ with $k\geq s$, then $p_s\dvds 2^n-1$, so $12\dvds n$ and hence $3^2\dvds 2^n-1$, which is impossible. Thus only the cases with $k<s$ are left.