Is a pair of bases $(a,b)$ known such that Wieferich primes to bases $a$ and $b$ are impossible? A prime number $p$ is called a Wieferich-prime to base $a$ if $$a^{p-1}\equiv 1\mod p^2$$ holds.

Is a pair $(a,b)$ known for which we know that a Wieferich-prime to base $a$ and $b$ is impossible ?

Probably for every base $a$, there are infinite many Wieferich-primes to base $a$, but maybe for primes simultaneously Wieferich to $2$ distinct bases, we have conditions that rule out such a prime for some pair.
 A: As far as I'm aware this problem is open. Interest in versions of this problem probably date to Mirimanoff in 1910 or shortly thereafter. Recall, Weiferich's original interest:
Theorem (Weiferich): If there is an integer solution to $$x^p+y^p=z^p$$ for an odd prime and $(xzy,p)=1$ then $2^{p-1} \equiv 1$ (mod $p^2$).  
Note that Weiferich's theorem does not by itself imply that any counterexample to Fermat's Last Theorem must be a Wieferich prime, since we required the additional criterion that $(xyz,p)=1$ which is sometimes called the "First case of Fermat's Last Theorem." In practice, prior to Wiles this was a pretty common additional assumption to make since it makes it much easier to prove things.
Following Weiferich, we then had:
Theorem: (Mrimanoff 1910): If there is an integer solution to $$x^p+y^p=z^p$$ for an odd prime and $(xzy,p)=1$ then $3^{p-1} \equiv 1$ (mod $p^2$). 
So, whether there's any such prime which satisfies both congruences became a natural question. Later work in a series of papers by Pollaczek and Granville and others they showed the same result with the 3 replaced by any prime which is less than or equal to 89. Which is pretty darn restrictive. 
Note that the sort of congruence under discussion was studied before Weiferich, with Abel and Jacobi both aware of some small similar examples such as $3^{10} \equiv 1$ (mod $11^2$) and $14^{28} \equiv 1$ (mod $29^2$). For all of the above, with references, see the section on Weiferich primes in Ribenboim's "The New Book of Prime Number Records" although he doesn't explicitly discuss your question. 
There are heuristic reasons for strongly suspecting that for any almost any two primes $a$ and $b$ one chooses, the answer to your question will be yes, and that even for others there should be only finitely many. Here's the basic argument: Define $$q(a,p)= \frac{a^{p-1}-1}{p}.$$
Of course by Fermat's Little Theorem, $q(a,p)$ is an integer whenever $(a,p)$ are relatively prime. Now, the "chance" that $q(a,p) \equiv 0$ (mod p) should be about $1/p$, and this is equivalent to $p$ being $a$-Weiferich. Now, the chance then that $q(a,p) \equiv 0$ (mod $p$) and $q(b,p)$ should be about $(1/p)(1/p)$. Now the expected number of such primes is then roughly $$\sum_{p} \frac{1}{p^2}$$ which is a quickly converging series. So we should expect only finitely many such primes. 
We can of course, construct examples of $a$ and $b$ which do share such a prime, simply by starting with $a \equiv b \equiv 1$ (mod $p^2$) for whatever your preferred odd prime $p$ is. I'd strongly suspect that aside from possibly a few obviously silly cases like this, there won't be any such primes $p$ for any choices of primes $a$ and $b$. Giving that we cannot even prove that there are no such solutions where instead of just $a$ and $b$ one has every prime from $3$ to $89$ as the bases, proving substantial results about anything like this seems very hard. 
