No integral solutions to y cubed equals x squared plus twenty-two ($y^3=x^2+22$)

The equation $y^3=x^2+22$ has no integer solutions. Factoring it in $\mathbb{Z}[\sqrt{-22}]$, the two conjugate factors are coprime. Unique factorization of ideals in a number field and the absence of three-torsion in the class group force $x+\sqrt{-22}$ to be a cube up to a unit. Writing it as $(a+b\sqrt{-22}{)}^3$ makes its $\sqrt{-22}$ coefficientwhich has no integer solution.