This is a supplement to Lucia's answer. His/her heuristic analysis suggests that there are infinitely many odd $m$'s such that the largest square dividing $m^3-2$ exceeds $9m^2$, contradicting the conjecture.
On the other hand, the abc conjecture implies that the largest square dividing $m^3-2$ is $\ll_\epsilon m^{2+\epsilon}$ for any $\epsilon>0$, which indicates that a slightly stronger version of the conjecture is probably true.
Some calculations with SAGE show that for $m<2\cdot 10^6$ there are only two instances where the largest square dividing $m^3-2$ exceeds $m^2$, namely $m=3$ and $m=100$.