The only solutions of $x^3-2y^3=1$ are $(x,y)=(1,0)$ and $(x,y)=(-1,-1)$. I don't know whether there's a nice Skolem-style proof, but here this happens to be unnecessary because $(1,0)$ and $(-1,-1)$ are the only rational solutions and this can be proved by a Fermat-style descent: the Weierstrass form is $Y^2 = X^3 - 27$, and there's a $2$-torsion point at $(X,Y)=(3,0)$. One could also use descent via $3$-isogeny to $Y^2 = X^3+1$, which has $6$ rational points, at $\infty$, $(-1,0)$, $(0,\pm1)$, and $(2,\pm3)$.
ADDED LATER: 1) As I already reported in a comment, the result on $x^3 - 2y^3 = 1$ turns out to be due to Euler himself. I found the reference in Dickson's History of the Theory of Numbers, Vol II on page 572: it is Theorem 247 in Euler's Elements of Algebra, see p.456 ff. of this English translation (Google Books scan of a Harvard library book from 1829). It looks like Euler chose to use a 3-descent (presumably because it was in the context of equations of the form $ax^3+bx^2+cx+d = y^3$), even though a 2-descent was also available.
- Meanwhile Rene Schoof notes that his book Catalan's Conjecture reproduces a 3-adic proof using Skolem's method, "from Bill McCallum's 1977 honours project at the University of Sydney". See Proposition 4.1, pages 17-19. [The $\root 3 \of 4$ coefficient of $(\root 3 \of 2 - 1)^n$ is $0 \bmod 3$ iff $n = 3k$ or $n = 3k+1$, and in both cases it vanishes mod $3^e$ iff $k$ does (each $e=1,2,3,\ldots$, by induction on $e$), whence the known zeros for $k=0$ are the only ones.] In the first paragraph of page 17, Schoof cites Euler's proof by descent, which he gives later in the book in an Appendix.