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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.

For nonzero ideals, unique factorization of ideals in a number field converts $AB=C^3$ to $v_P(A)+v_P(B)=3v_P(C)$. Coprimality says at most one of the first two exponents is positive, so each is divisible by three. Dividing its exponents by three constructs ideals whose cubes are $A,B$. If a zero ideal is allowed, coprimality forces the other ideal to be the whole ring and the conclusion is immediate with the zero ideal as its own cube.

To compute the Ideal class group of Q of square root minus thirty, put $\theta =\sqrt{-30}$. The ring of integers is $\mathbb{Z}[\theta ]$, with discriminant $-120$. The given Minkowski bound for ideal classes is $2\sqrt{120}/\pi =4\sqrt{30}/\pi <7$. Hence the ideal class group is generated by prime ideals with ideal norm below seven. The rational primes two, three and five all ramify, giving unique prime ideals $P_2=(2,\theta )$, $P_3=(3,\theta )$, $P_5=(5,\theta )$ with $P_i^2=(i)$ and ideal norm $i$. There is no additional prime ideal of ideal norm four because two ramifies with residue degree one; six is not a prime power. Thus these three classes generate and each has order at most two.

As ${\theta }^2=-30$, unique factorization of ideals in a number field gives $(\theta )=P_2{P}_3{P}_5$, so their three classes multiply to one. None is principal: a generator of one would be an integral element $a+b\theta$ of field norm $a^2+30b^2$ equal to two, three or five, and none of those field norm equations has integer solutions. The three nontrivial classes are therefore distinct, with the third the product of the first two. Hence

To rule out integer solutions, suppose $x^2+30=y^3$. Necessarily $y>0$. If a prime $p\mid 30$ divided $y$, then reduction modulo $p$ would force $p\mid x$. Because thirty is squarefree, the valuation of $x^2+30$ at that prime would be exactly one, while that of $y^3$ is a positive multiple of three, a contradiction. Thus $\gcd (y,30)=1$.

If a prime ideal divided both $(x+\theta )$ and $(x-\theta )$, it would divide their difference $(2\theta )$ and their product $(y{)}^3$. Its underlying rational prime would divide $N(2\theta )=120$, hence divide thirty, and would also divide $y$. This contradicts the gcd just proved, so the two ideals are coprime. Their product is $(y{)}^3$, so the first part gives $(x+\theta )=A^3$ for an ideal $A$.

In a ideal class group of exponent two, $[A{]}^3=1$ forces $[A]=1$. Thus $A=(a+b\theta )$ with $a,b\in \mathbb{Z}$. Equality of principal ideals gives $x+\theta =\epsilon (a+b\theta {)}^3$ for a unit $\epsilon$. The only units are $\pm 1$, since $a^2+30b^2=1$ forces $b=0$, $a=\pm 1$; both signs are cubes and can be absorbed into the generator. Comparing the coefficient of $\theta$ after cubing now giveswhich is impossible modulo three. Therefore