Recall that a *Wieferich prime* is a prime number $p$ such that
$2^{p-1} \equiv 1 \bmod p^2.$
It is not known whether there are infinitely many Wieferich primes, nor whether there are infinitely many non-Wieferich primes. In fact there are only $2$ known Wieferich primes.


I'm interested in a slightly different condition which I'm hoping is easier to handle. Namely, I can replace the exponent $p-1$ by the order of $2$ in $(\mathbb{Z}/p\mathbb{Z})^\times$. Moreover, I just want this power to hit the identity with odd $p$-adic valuation. Specifically:



> 
> Are there infinitely many primes $p$ such that $v\_p(2^{\mathrm{ord}\_p(2)}-1)$ is odd?
> 
> 
> 


Here $\mathrm{ord}\_p(n)$ denotes the order of $n$ in $(\mathbb{Z}/p\mathbb{Z})^\times$ (which divides $p-1$ by FLT), and $v\_p$ is the $p$-adic valuation.


Note that the existence of infinitely many non-Wieferich primes would provide a positive answer to my question (since here $v\_p(2^{p-1}-1) = 1$).


Ideally I'd also like to know that there collection of such primes has positive density, rather than just being infinite.



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.)


