= Solution
For <integral solutions of x squared minus two equals y cubed>, the equation implies $y\ge-1$, so $y+2>0$. If $y$ were even, then $x$ would be even and $x^2-2$ would be two or six modulo eight, whereas $y^3$ would be zero modulo eight. Thus $y$ and $x$ are odd. Also $3\nmid x$: otherwise $x^2-2\equiv7\pmod9$, which is not a cubic residue modulo nine. Reducing modulo three then gives $y\equiv2\pmod3$. Consequently
$$
\boxed{\gcd(y+2,6)=1.}
$$
Put $\eta=\sqrt{-6}$ and $\alpha=x+(y+1)\eta$. Its <field norm> is $(y+2)^3$. The <principal ideals> $(\alpha)$ and $(\bar\alpha)$ are coprime. Indeed a common <prime ideal> has underlying rational prime $p$ dividing $y+2$. The displayed gcd excludes two and three, so both two and $\eta$ are invertible modulo that <prime ideal>. The difference $\alpha-\bar\alpha=2(y+1)\eta$ therefore implies $p\mid y+1$, contradicting $p\mid y+2$.
Their product is $(y+2)^3$, so <unique factorization of ideals in a number field> and coprimality imply $(\alpha)=A^3$ for an <integral ideal> $A$. In the given <ideal class group> of order two, $[A]^3=1$ implies $[A]=1$. Hence $A=(u+v\eta)$ with <integers> $u,v$. Equality of <principal ideals> gives $\alpha=\varepsilon(u+v\eta)^3$. The only <units> are $\pm1$, and both are cubes, so the sign can be absorbed into $u,v$. Taking <norms>, and using $y+2>0$, gives
$$
\boxed{x+(y+1)\eta=(u+v\eta)^3,\qquad u^2+6v^2=y+2.}
$$
The coefficient of $\eta$ gives $y+1=3u^2v-6v^3$. Eliminating $y$ with the <field norm> identity yields
$$
(3v-1)u^2=6v^3+6v^2-1,\qquad u^2=\frac{6v^3+6v^2-1}{3v-1}.
$$
The denominator is nonzero for integral $v$. <Polynomial> division makes the integrality obstruction especially clear:
$$
\boxed{9u^2=18v^2+24v+8-\frac1{3v-1}.}
$$
If $u$ is integral, the reciprocal must be an <integer>, forcing $3v-1=\pm1$. The only integral possibility is $v=0$. Then $u^2=1$, the <field norm> gives $y=-1$, and the real coefficient in the cube gives $x=u^3=\pm1$. Both pairs directly satisfy the original equation. Therefore all integral solutions are
$$
\boxed{(x,y)=(1,-1),\ (-1,-1).}
$$
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