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$\let\dvds\mid$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.