OPEN

Is there an integer $m$ with $(m,6)=1$ such that none of $2^k3^\ell m+1$ are prime, for any $k,\ell\geq 0$?

There are odd integers $m$ such that none of $2^km+1$ are prime, which arise from covering systems (i.e. one shows that there exists some $n$ such that $(2^km+1,n)>1$ for all $k\geq 1$). Erdős and Graham also ask whether there is such an $m$ where a covering system is not 'responsible'. The answer is probably no since otherwise this would imply there are infinitely many Fermat primes of the form $2^{2^t}+1$.