Since $-65 ≡ 3 \bmod 4$, we have $𝒪_K=ℤ[α]$ for $α=\sqrt{-65}$. The Minkowski constant is $c_K=(4 / π)(1 / 2) \sqrt{4 × 65}=(4 / π) \sqrt{65}<32.4 / π<11$. So we need to factorise $2,3,5,7$ using Dedekind's theorem.
$x^2+65=(x+1)^2$ so $(2)=P_2^2$ for $P_2=(2, α+1)$.
$x^2+65=x^2-1=(x-1)(x+1) \bmod 3$ so that $(3)=P_3 Q_3$ for $P_3=(3, α-1), Q_3=(3, α+1)$.
$x^2+65=x^2 \bmod 5$ so that $(5)=P_5^2$ for $P_5=(5, α)$.
$x^2+65=x^2-5 \bmod 7$ which is irreducible. Thus $(7)$ is a prime in $ℤ[α]$.
Therefore, the class group is generated by $[P_2],[P_3]$ and $[P_5]$. However, $N(5-α)=25+65=90=3^2 × 2 × 5$. Note that if $α-5$ were in $P_3$, then $(α-1)-3-(α-5)=1$ would be in $P_3$. Thus, $P_3 ∤(α-5)$, and we get the prime decomposition $(α-5)=Q_3^2 P_2 P_5$. Hence, $[P_5]=[P_2][P_3]^2$, and the class group is generated by $[P_2]$ and $[P_3]$. The element $[P_2]$ is of order 2. The only element $a+b α$ of norm $9(=a^2+65 b^2)$ is $± 3$. If we had $(3)=P_3^2$, then we would have $P_3=Q_3$. So $P_3^2$ is not principal. On the other hand, $N(4+α)=16+65=81$ and $4+α ∉ P_3$. (If it were, then $4+α-(α-1+3)=2$ would be in $P_3$ so that $P_3$ would divide (2), a contradiction.) Hence, we must have $(4+α)=Q_3^4$. Thus, both $[Q_3]$ and $[P_3]=[Q_3]^{-1}$ have order 4. If we had $[P_3]^2=[P_2]$, then $Q_3^2 P_2$ would be principal. Thus it would be generated by an element of norm 18, which doesn't exist. Hence, $[P_3]^2 ≠[P_2]$, so that $[P_2]$ is not in the subgroup generated by $[P_3]$. So the subgroups $[P_2]$ and $[P_3]$ intersect trivially, and we conclude that the class group is isomorphic to $C_4 × C_2$.

Let $(x, y)$ be an integral solution to $y^2=x^3-65$. Note that both $x$ and $y$ are coprime to $65=13 × 5$, since if either were divisible by 13 or 5, so would the other be. But then 65 would be divisible by $5^2$ or $13^2$. Also, if $x$ were even, then $x^3 ≡ 0 \bmod 8$. But then, $y^2 ≡-1 \bmod 8$, which is not possible. Hence, $x$ must be odd. Now write the equation as
\[
(y-\sqrt{-65})(y+\sqrt{-65})=x^3.
\]
Claim: $(y-\sqrt{-65})$ and $(y+\sqrt{-65})$ are coprime. Suppose a prime ideal $P$ divided both. Then it would divide $2 \sqrt{-65}$. It would also divide $x^3$. So $N(P)$ would divide $4 × 65$ and $x^6$. But $x$ is coprime to $4 × 65$.

From this, we conclude that all prime factors of $x$ divide only one of $(y-\sqrt{-65})$ and $(y+\sqrt{-65})$. Therefore, we have $(y+\sqrt{-65})=I^3$ for some ideal $I$. The ideal class $[I]$ has order dividing 3. But the class group has order 8. Hence, $[I]$ must be trivial, and $I=(α)$ is principal. Since the only solution to $x^2+65 y^2= ± 1$ is $x= ± 1, y=0$, we see that the only units in $𝒪_K$ are $± 1$. Therefore, after changing signs if necessary, we get $y+\sqrt{-65}=α^3$. Write $α=a+b \sqrt{-65}$. Then \[ α^3=(a+b \sqrt{-65})^3=a^3-195 a b^2+(3 a^2 b-65 b^3) \sqrt{-65}. \] So we get $(3 a^2-65 b^2) b=1$, By considering the two possibilities $b= ± 1$, we see right away that there are no solutions. Therefore, the original equation $y^2=x^3-65$ has no integral solutions.

From this, we conclude that all prime factors of $x$ divide only one of $(y-\sqrt{-65})$ and $(y+\sqrt{-65})$. Therefore, we have $(y+\sqrt{-65})=I^3$ for some ideal $I$. The ideal class $[I]$ has order dividing 3. But the class group has order 8. Hence, $[I]$ must be trivial, and $I=(α)$ is principal. Since the only solution to $x^2+65 y^2= ± 1$ is $x= ± 1, y=0$, we see that the only units in $𝒪_K$ are $± 1$. Therefore, after changing signs if necessary, we get $y+\sqrt{-65}=α^3$. Write $α=a+b \sqrt{-65}$. Then \[ α^3=(a+b \sqrt{-65})^3=a^3-195 a b^2+(3 a^2 b-65 b^3) \sqrt{-65}. \] So we get $(3 a^2-65 b^2) b=1$, By considering the two possibilities $b= ± 1$, we see right away that there are no solutions. Therefore, the original equation $y^2=x^3-65$ has no integral solutions.

It suffices to find $K$ whose class group has an element of order 2 and an element of order 3. If $K=ℚ(\sqrt{d})$ with $d<-1$ and $d ≡ 2,3 \bmod 4$, then $x^2-d ≡ x^2 \bmod 2$ or $x^2-d ≡(x+1)^2 \bmod 2$. Hence, we will have $(2)=P_2^2$ for an ideal $P_2$ of norm 2. On the other hand, $N(a+b \sqrt{d})=a^2-b^2 d$, so there is no element of norm 2. Thus, $P_2$ is not principal and has order 2.

Therefore, we need only choose $d$ so that we also get an element of order 3. We look for such an element of the form $P_l$ where $N(P_l)=l$ is a prime. Then we want $P_l^3$ to be principal, and hence, $l^3$ to be a norm. So we wish to be able to solve $l^3=a^2-b^2 d$. Since we've already arranged for divisors of 2 to have order 2, the smallest $l$ to try is 3. Thus, we get $3^3=a^2-b^2 d$. Trying the simplest case first, we could try this for $a=1, b=1$, in which case, we get $d=-26$. We have $x^2+26=x^2+2=x^2-1=(x-1)(x+1) \bmod 3$. Hence, $(3)=P_3 Q_3$ where $P_3=(3, \sqrt{-26}+1)$, and $Q_3=(3, \sqrt{-26}-1)$. If we had $1+\sqrt{-26} ∈ Q_3$, then $2 ∈ Q_3$ so $Q_3 ∣(2)$, a contradiction. Hence, the prime factorisation of $\sqrt{-26}+1$ is $P_3^3$. Therefore, $[P_3]$ has order 1 or 3. But there is no solution to $x^2+26 y^2=3$, and hence, $P_3$ is not principal. Therefore, $[P_3]$ is of order 3, and $[P_2][P_3]$ has order 6.

Therefore, we need only choose $d$ so that we also get an element of order 3. We look for such an element of the form $P_l$ where $N(P_l)=l$ is a prime. Then we want $P_l^3$ to be principal, and hence, $l^3$ to be a norm. So we wish to be able to solve $l^3=a^2-b^2 d$. Since we've already arranged for divisors of 2 to have order 2, the smallest $l$ to try is 3. Thus, we get $3^3=a^2-b^2 d$. Trying the simplest case first, we could try this for $a=1, b=1$, in which case, we get $d=-26$. We have $x^2+26=x^2+2=x^2-1=(x-1)(x+1) \bmod 3$. Hence, $(3)=P_3 Q_3$ where $P_3=(3, \sqrt{-26}+1)$, and $Q_3=(3, \sqrt{-26}-1)$. If we had $1+\sqrt{-26} ∈ Q_3$, then $2 ∈ Q_3$ so $Q_3 ∣(2)$, a contradiction. Hence, the prime factorisation of $\sqrt{-26}+1$ is $P_3^3$. Therefore, $[P_3]$ has order 1 or 3. But there is no solution to $x^2+26 y^2=3$, and hence, $P_3$ is not principal. Therefore, $[P_3]$ is of order 3, and $[P_2][P_3]$ has order 6.