If a CAS can do it, I guess a hand can do it, although my hands don't have much experience in the matter. Probably you do some kind of descent/ Selmer group calculation?

@pablo I just mean that the CAS can compute whole Mordell-Weil group over of the elliptic curve $\mathbb{Q}$. Apparently the group only has 3 points. To figure out what the torsion looks like by hand, you'd have to do some work, but these techniques are well-known.

Maybe it's worth pointing out that, even if one did not realize it was isomorphic to the Fermat curve, one could discover quickly using a CAS that the elliptic curve has rank 0, and then that the only rational points were the obvious ones (which I guess are 3-torsion points).

Over what field?

@MattF. I thought of the same example (basically) after posting this, but haven't had a chance until now to come and fix the question to avoid this. Take a look at the edit I am about to make.

Thanks for the references! I would accept both answers if I could. I accepted Joe's answer because he spoke to why one can't get better than 1/2 power savings.

Thanks for the examples, and thanks especially for noticing that I stated the opposite of what I meant. I have edited the question accordingly.

@FanZheng yes. Here is one effective result: ams.org/journals/mcom/1996-65-213/S0025-5718-96-00664-3/… edit: and here is the one I was probably looking at yesterday, due to Wu and Mu (Quanwu Mu, not the OP!): sciencedirect.com/science/article/pii/S0022314X12001989

Interesting question. I assume you're familiar with work of Smyth, Flammang, etc. on lower bounds for the Mahler measure of totally positive algebraic integers? (These give a lower bound of $C=1.722...$ for $M(\alpha)^{1/d}$ with finitely many exceptions, but valid for all totally positive integers $\alpha$, not just those with all conjugates $>1$).