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 $\sqrt{27}$. 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 $2d^{1/3}$, 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.