I've tried to prove it in contrapositive $\forall x \in \mathbb{Z}^+$ ($\neg$($a$ does not divide $2a^2+a+7$) $\rightarrow$ $\neg (a$ is a even number)), which basically means that if a divide the equation then a is an odd number. So:

$a$ is odd: $a = 2k+1$ for an arbitrary positive integer k.

$2(2k+1)^2+(2k+1)+1 = (2k+1)x$, since if $a$ can divide the equation, there must be another integer x such that $ax=2a^2+a+7$

But i don't know what to do next or even I'm on the right track.