Primitive integer solutions to $2x^2+xy+3y^2=z^3$? The class number of $\mathbb{Q}(\sqrt{-23})$ is $3$, and the form
$$2x^2 + xy + 3y^2 = z^3$$
is one of the two reduced non-principal forms with discriminant $-23$.
There are the obvious non-primitive solutions $(a(2a^2+ab+3b^2),b(2a^2+ab+3b^2), (2a^2+ab+3b^2))$.
I'm pretty sure there aren't any primitive solutions, but can't seem to find an easy argument. Are there?
In general, is it possible for a non-principal form to represent an $h$-th power primitively (where $h$ is the class number of the associated quadratic field)?

[EDIT]

I think I've solved the first question and have posted an answer below.
Since the proof is very technical I don't see how it can generalize to the greater question above ($h$-th powers).

If $4|y$, then $2|z$, which in turn, by reducing mod $4$, $2|x$, contradicting primitivity. Multiply by $2$ and factor the equation over $\mathbb{Q}(\sqrt{-23})$:
$$(\frac{4x+y+\sqrt{-23}y}{2})(\frac{4x+y-\sqrt{-23}y}{2}) = 2z^3$$
Note that both fractions are integral. The gcd of the two factors divides $\sqrt{-23}y$ and $4x+y$. If $23|4x+y$ then $23|z$, and reducing the original equation modulo $23^2$ we see that $23|y$, hence also $23|x$, contradicting primitivity. So the gcd divides $y$ and $4x+y$, and by the above argument, it then must be either $1$ or $2$ according to wether $y$ is odd or even, respectively.
First, assume that $y$ is odd. So the gcd is $1$. Thus, for some ideal $I$:
$$(2x+y\frac{1+\sqrt{-23}}{2})=(2,\frac{1+\sqrt{-23}}{2})I^3$$
Which implies that $(2,\frac{1+\sqrt{-23}}{2})$ is principal, leading to a contradiction. Now assume that $y$ is even, and that $x$ is therefore odd and $z$ is even. Put $y=2u$, $z=2v$, $u$ odd, so that:
$$(2x+u+\sqrt{-23}u)(2x+u-\sqrt{-23}u)=16v^3$$
Both factors are divisble by $2$, so that:
$$(x+u\frac{1+\sqrt{-23}}{2})(x+u\frac{1-\sqrt{-23}}{2})=4v^3$$
As before, the gcd is 1, and since $x+u\frac{1+\sqrt{-23}}{2}=x+u+u\frac{-1+\sqrt{-23}}{2} \in (2, \frac{-1+\sqrt{-23}}{2})$ we must have for some ideal $I$:
$$(x+u\frac{1+\sqrt{-23}}{2}) = (2, \frac{-1+\sqrt{-23}}{2})^2I^3$$
Contradicting that $(2, \frac{-1+\sqrt{-23}}{2})^2$ is non-principal (the ideal above $2$ appears squared since the product of factors is divisible by $4$). We are done!

It actually seems that the above can be generalised, but I have to use a major theorem, which seems like it might be a bit of an overkill. Without further ado:
Let $aX^2+bXY+cY^2$ be a primitive non-principal quadratic form of discriminant $\Delta=b^2-4ac$, and $h$ be the class number of the associated quadratic field. Assume there is a solution $x,y,z$:
$$ax^2+bxy+cy^2=z^h$$
Recalling the Chebotarev Density Theorem (OVERKILL), there is an equivalent form with $a$ an odd prime that doesn't divide $\Delta$, and since the invertible change of variables preserves primitivity, we reduce to this case.
Multiplying by $a$ and factoring over $\mathbb{Q}(\sqrt{\Delta})$:
$$(ax+\frac{b+\sqrt{\Delta}}{2}y)(ax+\frac{b-\sqrt{\Delta}}{2}y)=az^h$$
The gcd of the factors divides $(2ax+by,\sqrt{\Delta}y)$. Say $\Delta |2ax+by$, then since
$$(2ax+by)^2-\Delta y^2=4az^h$$
we must have $\Delta |z$, so $\Delta |y$, and finally $\Delta |x$ (unless $\Delta=\pm 2$, which is impossible), contradicting primitivity.
Hence the gcd divides $(2a,y)$.
1) gcd$=1$:
$$(ax+\frac{b+\sqrt{\Delta}}{2}y) = (a,\frac{b+\sqrt{\Delta}}{2})I^h$$
contradicting that the form is non-principal.
2) gcd$=2$ or $2a$: then $y$ is even, so $z$ is too, and since $a$ is odd, $x$ must also be even, contradicting primitivity.
3) gcd$=a$: thus $a|z$. If $a^2|y$, then reducing modulo $a$ we see that $a|x$. So $a||y$. Dividing this gcd out of the two factors, we must have:
$$(x+\frac{b+\sqrt{\Delta}}{2}\frac{y}{a}) = (a,\frac{b-\sqrt{\Delta}}{2})^{h-1}I^h$$
for some ideal $I$. Note the particular ideal above $a$, which appears since its conjugate cannot appear as $x$ isn't divisble by $a$. Since $(a,\frac{b-\sqrt{\Delta}}{2})^{h-1} \sim (a,\frac{b+\sqrt{\Delta}}{2})$, this contradicts that the form is non-principal.
We are done!
There must be a way to avoid applying a major theorem such as Chebotarev... Please tell me if you have an idea how :D
