A variant on Wieferich primes

Suppose the answer is no and that the finitely many exceptions are all at most $B$. Let $\ell \equiv 1 \pmod{3}$ be prime and consider $n=2^{\ell} -1$. If $p>B$ is a factor of $n$, then $\ell$ is the order of $2$ modulo $p$, so $p$ occurs in $n$ with an even exponent, so $n = x^2c, c \le B!$. Let $y = 2^{(\ell - 1)/3}$. Then $n=2y^3 - 1$ and finally $2y^3 - 1 = cx^2$, so $(x,y)$ is an integral point on one of a finite collection of elliptic curves and there can be only finitely many such. But there are infinitely choices for $\ell$, contradiction. (This is a variant of an old argument of Granville.)

Felipe refers to my first ever paper (mathscinet.ams.org/mathscinet-getitem?mr=789713) from 1985 ! However I have a more recent paper that gives a better result along the lines asked for (mathscinet.ams.org/mathscinet-getitem?mr=2997580) which shows that every $2^n-1$, with $n\ne1$ or $6$, has a primitive prime factor that divides it to an odd power.