No integer solutions to y cubed equals x squared plus thirty ($y^3=x^2+30$)

The equation has no integer solutions. Work in the ring of integers $\mathbb{Z}[\theta ]$ for $\theta =\sqrt{-30}$. If a rational prime divides both $y$ and thirty, it divides $x$, and the valuation of $x^2+30$ is one because thirty is squarefree, contradicting the valuation of a cube. Thus $\gcd (y,30)=1$. The two principal ideals $(x+\theta ),(x-\theta )$ are coprime: any common prime ideal lies above a prime dividing both $N(2\theta )=120$ and $y$. Their product is $(y{)}^3$, so unique prime ideal factorization gives $(x+\theta )=A^3$. The Ideal class group of Q of square root minus thirty has exponent two, forcing $A$ to be principal. The only units are $\pm 1$ and both are cubes, so $x+\theta =(a+b\theta {)}^3$ for integers $a,b$. Comparison of the coefficient of $\theta$ gives $1=3b(a^2-10b^2)$, impossible modulo three.