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
<a href="https://books.google.com/books?id=X8yv0sj4_1YC&dq=euler+elements+of+algebra&printsec=frontcover&source=bn&hl=en&ei=KiwwS8bdEcutlAe2waCoBw&sa=X&q=247">this
English translation</a> (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.

2) 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.