Wieferich prime-Lang-Trotter conjecture connection? Crandall-Dilcher-Pomerance prediction states that the number of Wieferich primes $&lt;x$ is $log\ logx $
N.Katz in "WIEFERICH PAST AND FUTURE" states;

The Crandall-Dilcher-Pomerance prediction is analogous to the Lang-Trotter
conjecture [L-T] for an elliptic curve $E$ over $Q$.

So my question is; if it is proven that there are only 2 Wieferich primes base 2, does that mean the Lang-Trotter conjecture is false?
I digged in the reference:The reference is to Serge Lang, Old and new conjectured diophantine inequalities, Bulletin of the American Mathematical Society, 23(1):37–75, 1990. 

On page 42 it is stated:

Actually, in line with Lang-Trotter conjectures, the probability
that $2^{p-1} = 1 + pk\pmod{p^2}$ with a fixed residue class $k \pmod p$
should be $O(1/p)$, so the number of primes $p\leq x$ such that
$2^{p-1} = 1 \pmod {p^2}$ should be $O(\log\log x)$.

So, I translate this as: "If there are only 2 Wieferich primes base 2, the Lang-Trotter conjecture is false". Any comments?
I digged in the reference:The reference is to Serge Lang, Old and new conjectured diophantine inequalities, Bulletin of the American Mathematical Society, 23(1):37–75, 1990. 

On page 42 it is stated:

Actually, in line with Lang-Trotter conjectures, the probability
that $2^{p-1} = 1 + pk\pmod{p^2}$ with a fixed residue class $k \pmod p$
should be $O(1/p)$, so the number of primes $p\leq x$ such that
$2^{p-1} = 1 \pmod {p^2}$ should be $O(\log\log x)$.

So, I translate this as: "If there are only 2 Wieferich primes base 2, the Lang-Trotter conjecture is false". Any comments?