Integral solutions of x squared minus two equals y cubed ($x^2-2=y^3⟹(x,y)=(\pm 1,-1)$)

In the ring of integers $\mathbb{Z}[\sqrt{-6}]$, which has class number two and units $\pm 1$, write $\alpha =x+(y+1)\sqrt{-6}$. Its norm is $(y+2{)}^3$. Parity modulo eight and cubic residues modulo nine show that $y$ is odd and $3$ does not divide $y+2$, so $\gcd (y+2,6)=1$. The ideals of $\alpha$ and its conjugate are coprime: a common prime cannot lie above two or three by their norms; at any other prime their difference forces divisibility of $y+1$, contradicting divisibility of $y+2$.

Unique factorization of ideals gives $(\alpha )=A^3$. The class group of order two forces $A$ principal, and its unit factor can be absorbed into a cube. Hence $\alpha =(u+v\sqrt{-6}{)}^3$ with integers $u,v$, and $u^2+6v^2=y+2$. Comparing coefficients and eliminating $y$ yieldsIntegrality requires $3v-1=\pm 1$, so $v=0$ is the only integral possibility. Then $u=\pm 1$, $y=-1$, and $x=\pm 1$. Direct substitution verifies both solutions. The class-group step is what permits extracting a cube of an element from a cube of an ideal despite failure of unique factorization of elements.