Do there exist super-Wieferich primes? A Wieferich prime is a prime $p$ such that $2^{p-1}\equiv 1\mod{p^2}$.
Denote the order of $2$ modulo $p$ by $O(p)$. Then we can show that a prime $p$ is a Wieferich prime if and only $O(p^2)=O(p)$. Otherwise we have $O(p^2)=pO(p)$.


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*What do we know of $O(p^3)$ for Wieferich primes?

*Do there exists super-Wieferich primes, that is, primes $p$ such that $O(p^3)=O(p)$.

 A: We only know of two Wieferich primes $ 1093$ and $3511$, and they are not super-Wieferich.
Naively we should expect that a "random" prime $p$ has probability $1/p^2$ of $2^p \equiv 2 \mod p^3$ (since we know $2^p \equiv 2 \mod p$, there are $p^2$ possible values for $2^p \mod p^3$).  Since $\sum_p 1/p^2 < \infty$, we should expect that there are only finitely many super-Wieferich primes.  Since we've tested lots of primes in the search for more Wieferich primes (up to at least $4.97 \times 10^{17}$) and the sum of $1/p^2$ for those we haven't tested is very small, I am reasonably confident that there are no super-Wieferich primes.  Of course this is not anywhere near a proof. 
On the other hand, since $\sum_p 1/p = \infty$ we expect there to be infinitely many Wieferich primes.
EDIT: To support the idea that $2^p - 2 \mod p^3$ should be a random multiple of $p$, here is a plot.  Each black dot $(x,y)$ is obtained as follows:
if $p$ is the $x$'th odd prime, for $1 \le x \le 2000$, $y = z/p^3$ where $0 \le z < p^3$ and $2^p - p \equiv z \mod p^3$.  It looks pretty random to me. But, as I said, this is heuristic.

