Without solving it, is there an elementary way to show that $X^3+Y^3=Z^3$ has a finite number of primitive [and non-trivial] integer solutions?

Considering the cubic case of Fermat’s Last Theorem, I make the following claim:

Proposition: The Diophantine equation $$X^3 + Y^3 = Z^3 \tag{\star}$$ has a finite number of primitive [and non-trivial] integer solutions $(x,y,z)$.

Is there a simple and elementary way to prove this statement?

Note: I am aware of Wiles’s proof of FLT in the general case, and the infinite descent proof of the cubic case by Euler et al., and the Mordell–Weil theorem, etc. I’m just curious, independent of those things, whether this weaker proposition has a simple and elementary proof.

• So you mean really different, since if $x,y,z$ is a solution so is $(ax,ay,az)$. – André Nicolas Mar 27 '16 at 19:32
• You never know. You should be aware of Selmer's example, $3x^3 + 4 y^3 + 5 z^3.$ There are no local obstructions to this being zero, however it cannot be zero with integers $x,y,z$ unless all three are zero. math.uconn.edu/~kconrad/blurbs/gradnumthy/selmerexample.pdf – Will Jagy Mar 27 '16 at 19:34
• – Watson Mar 27 '16 at 19:38
• @AndréNicolas: Thanks — I edited the question and Proposition accordingly. Hope it more clearly and accurately represents what I'm asking. – Kieren MacMillan Mar 27 '16 at 19:41
• @WillJagy: Nice! I look forward to absorbing that. – Kieren MacMillan Mar 27 '16 at 19:42