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 weaker version of the conjecture (i.e. one with a more restrictive constraint on the variables) is probably true.

Some calculations with SAGE show that for $m<10^7$ there are only two instances where the largest square dividing $m^3-2$ exceeds $m^2$, namely $m=3$ and $m=100$. (**Update:** Noam Elkies has extended the range to $m<10^8$, see his comments.) I found five further values with a square divisor exceeding $m^2/2$, and these are $\{1244,11317,296428,714417,722428\}$.

**Added.** The constant in Lucia's heuristic analysis (with $m$ restricted to odd numbers) equals
$$ \frac{1}{6}\frac{1}{2}\frac{2}{3}\prod_{\substack{p>3\\f(p)=0}}\left(1-\frac{1}{p^3}\right)\prod_{\substack{p>3\\f(p)=1}}\left(1-\frac{1}{p^2}\right)\prod_{\substack{p>3\\f(p)=3}}\left(1-\frac{3}{p^2}+\frac{2}{p^3}\right)$$
times the residue of the Dedekind zeta function of $\mathbb{Q}(\sqrt[3]2)$. The initial factor $1/6$ accounts for Lucia's denominator $3$ and the fact that we only look for odd $m$'s. The other factors come from comparing the Euler factors of $\sum_{\ell=1}^\infty f(\ell)\ell^{-s}$ to those of the Dedekind zeta function. My calculation in SAGE shows that
$ C\approx 0.423$, which suggests that the average ratio between the $\ell$'s yielding a counterexample $(\ell,m)$ is about $1.867\times 10^{10}$. 

I extended Noam Elkies' gp-pari calculation to the range $\ell<1.5\times 10^9$ (and continuing beyond). The best pair I found so far was $(\ell,m)=(228647269, 340601626)$ with a ratio $m/\ell\approx 1.4896$.