Is it true that there is an infinite number of prime numbers $p$ for which there exists at least one natural number $m_p$ such that $p$ is contained in the decimal expansion of $2^{m_p}$?
To clarify what I exactly mean, let us take as an example $p=19$. Then we can take $m_p=13$ because $2^{13}=8192$ and we see that $p=19$ is contained in the decimal expansion of $2^{13}$.
