If your students know a little about the Eisenstein integers (unique factorization and what the units are) there's the following simple argument (maybe it's essentially Euler's?).
Let u and v be the complex roots of z^2+z+1=0.

Theorem: Let A,B,C be non-zero elements of Q[u] with sum 0 and product twice a cube. Then
some two of A,B,C are equal.

Corollary: Suppose x is in Q and x^2-1 is a cube. Then x is 1,-1,0,3, or -3.

(To prove the corollary let A=1+x, B=1-x and C=-2).

The proof of the theorem is a reductio. If there's a counterexample, there's one with A,B,C in Z[u]; take such a counterexample with d=min(/A/,/B/,/C/) as small as possible. Then A,B and C are pairwise prime. Since ABC=2(cube) we may assume A=2i(cube), B=j(cube) C=k(cube) where i,j,and k are in the set {1,u,v}. Now all cubes in Z[u] are 0 or 1 mod 2. Since B+C is 0 mod 2, j=k. Since ABC=2(cube), ijk is a cube and i=j=k. We may assume i=j=k=1. Then A=2r^3,B=s^3,C=t^3, and we may further assume that s and t are 1 mod 2. s and t are not both in {1,-1} and it follows that d is at least root(27). Now look at s+t, us+vt and vs+ut. They
sum to 0 and their product is B+C=-2(r^3). They are congruent to 0,1 and 1 mod 2, and the last 2 of them can't be equal since s is not equal to t. Since each of them is at most 2(d^(1/3), this contradicts the minimality assumption.

This is really a 3-descent argument on an elliptic curve, but the fancy language as you see isn't needed.An almost identical argument gives what I think is the nicest proof of FLT for exponent 3.