Note that it suffices to only prove that for each prime number $p$ there are $a,b$ such that $m^3+am+b$ is never divisible by $p$. Equivalently, we should prove that there exists a number $a$ such that $P(m)=m^3+am$ is not surjective $\mod p$. Note though that $a=-7$ works indeed since $P(1)=P(2) \mod p$ and $2-1$ is not divisble by $p$. Hence we can choose a suitable $b$ as well and we are done.