$\DeclareMathOperator{\ord}{ord}$Artin's conjecture stipulates that $\ord_p(2) = p -1$ for infinitely many primes $p$, where $\ord_p(2)$ denotes the multiplicative order of $2$ modulo $p$. More generally one expects that $\ord_p(2)$ is often quite large. I'm looking for a weakened version of this, namely:
Does the sum $\displaystyle\sum_{p \leq x} \frac{1}{\ord_p(2)^2}$ converge as $x\to\infty$?
I would prefer unconditional results, but results conditional on e.g. GRH are still welcome.
Since Joe Silverman raised the possibility of variations of the question, I want to point out (as a long comment) that even a mild variation
$$ \sum_{p } \frac{1}{ \operatorname{ord}_p(2)^2 \log ( \operatorname{ord}_p(2))^\epsilon}$$
provably converges, since (following an argument given by Julian Rosen in the comments) $\operatorname{ord}_p(2)=d$ only if $p$ divides $2^d-1$, and the number of such is $$\omega(2^d-1) = O \left( \frac{\log(2^d-1)}{ \log (\log(2^d-1))} \right) = O \left( \frac{d}{ \log d} \right)$$
so
$$ \sum_{p } \frac{1}{ \operatorname{ord}_p(2)^2 \log ( \operatorname{ord}_p(2))^\epsilon} \leq \sum_d \frac{1}{ d^2 \log(d)^\epsilon} O \left( \frac{d}{ \log d} \right) = O \left( \sum_d \frac{1}{ d( \log d)^{1+\epsilon}}\right) <\infty.$$
One can try to improve this argument by taking advantage of the fact that one $p$ can't divide $2^d-1$ for too many different $d$s, but I don't think you will solve it with such reasoning. A bad scenario you would need to rule out is that for each $d$ there are $ \sim c_1 d / \log d$ primes, all of size $\sim d^{c_2}$, that divide $2^{d}-1$, for some arbitrary constants $c_1,c_2$ with $c_2 > 3$ and $c_1 c_2 < \log 2 $.
We want $c_2>3$ so the number of primes between $d^{c_2}$ and $(d+1)^{c_2}$ that are congruent to $1$ modulo $d$ is still at least $\sim d/\log d$, and we want $c_1 c_2 < \log 2$ so that the product of all these primes is still less than $2^{d}-1$.
Not quite what you've asked for, but in case it helps in whatever application you have in mind: $$ \sum_{p~\text{prime}} \frac{\log p}{p \operatorname{ord}_p(a)^\epsilon} \le \log\log a + \frac{2}{\epsilon} + C $$ for all $\epsilon>0$ and an absolute constant $C$. Here $a\in\mathbb Z$ is any integer with $|a|\ge2$. This will at least tell you that $\operatorname{ord}_p(a)$ cannot be too small, too often. The proof, which is fairly elementary, is in Variations on a theme of Romanoff, Internat. J. Math. 7 (1996), 373-391 [MR1395936].
