To add to Abhinav's answer: the fact that $x^3+y^3=2$ has no solutions
other then $x=y=1$ is attributed by Dickson to Euler himself:
see Dickson's *History of the Theory of Numbers* (1920) Vol.II, Chapter XXI
"Numbers the Sum of Two Rational Cubes", page 572. The reference
(footnote 182) is
"Algebra, 2, 170, Art. 247; French transl., 2, 1774, pp. 355–60;
Opera Omnia, (1), I, 491". In the next page Dickson also refers to
work of Legendre that includes this result (footnote 184: "Théorie
des nombres, Paris, 1798, 409; ...").

This result is actually easier than the $n=3$ case of Fermat,
because the curve has a $2$-torsion point, so only a $2$-descent
is required, and here one soon finds that there are only two
rational points (the second being the point at infinity
$(x:y:1) = (1:-1:0)$). [It also happens that this curve has
conductor $36$, while it's known that the smallest conductor
of an elliptic curve of positive rank is $37$; but that's
using a huge modern cannon to dispatch a classical fly.]

An easy way to bring the curve $x^3+y^3=2$ into Weierstrass form
is to factor $x^3+y^3 = (x+y)(x^2-xy+y^2)$ and substitute
$(x,y)=(r+s,r-s)$ to get $r^3+3rs^2=1$. Thus $s^2=(1-r^3)/(3r)$,
so $3r(1-r^3)$ is to be a square. Now substitute $r=3/X$ to get
$s^2 = (X^3-27)/(3X)^2$, or equivalently $Y^2=X^3-27$ where $y=3sX$.
The solution $(x,y)=(1,1)$ corresponds to $(r,s)=(1,0)$ and then
$(X,Y)=(3,0)$, which we recognize as a 2-torsion point because $Y=0$.