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he 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 $(2,\pm3)$.