Hodge numbers of a K3 surface over general field

Let $S$ be a K3 surface over some field $k$. That is, $S$ is a nice variety over $k$ such that the canonical bundle $\omega_S$ is trivial and $H^1(S, \mathcal{O}_S) = 0$. As an exercise for myself, I wanted to see if I can compute the Hodge numbers $h^{p,q} := \dim_k H^q(S,\Omega^p_S)$. I have computed the Hodge numbers for $p=0,2$ and found that $h^{1,0} = h^{1,2}$ and $h^{1,1} = 20 + 2 h^{1,0}$ (I will explain below how I did this). I know what the answer is supposed to be, so it is now sufficient to show that $h^{1,0} = 0$. In the case that $k= \mathbb{C}$, I am able to show this (I will also put the argument below). I was now wondering how one can finish this for general $k$?

I have seen online that K3 surfaces are hyperkähler and this gives rise to relations between the Hodge numbers, but this is something I would like to avoid using since I don't really know where it comes from or why a K3 surface should satisfy it. I also don't know if/how this hyperkähler business extends to fields other than $\mathbb{C}$, since what I have found online seems to phrased in terms of differential geometry.

How far I got with the Hodge numbers for general $k$: Since $S$ is projective and irreducible, we have that $h^{0,0} = 1$. Since $\omega_S = \mathcal{O}_S$, Serre duality then gives that $h^{0,2} = 1$. By the definition of $S$, we also have that $h^{0,1} = 0$. We certainly also have $h^{2,q} = h^{0,q}$ for each $q$. Next, using Hirzebruch-Riemann-Roch on $\mathcal{O}_S$ gives that

$$ 2 = h^{0,0} - h^{0,1} + h^{0,2} = \int_S ch(\mathcal{O}_S) Td_S = \frac{1}{12} \int_S c_2(T_S), $$ where $T_S$ is the tangent bundle of $S$. Here we have used that $c_1(T_S) = - c_1(\omega_s) = 0$. We conclude that $\int_S c_2(\Omega_S) = \int_S c_2(T_S) = 24$. Now using Hirzebruch-Riemann-Roch on $\Omega_S$ gives $$ h^{1,0} - h^{1,1} + h^{1,2} = \int_S ch(\Omega_S) Td_S = 4 - \int_S c_2(\Omega_S) = -20. $$ Finally, a rank 2 bundle is isomorphic to its dual up to tensoring by the determinant (Hartshorne Exercise II.5.16.b), so $\Omega_S = T_S$. Serre duality then gives that $h^{1,0} = h^{1,2}$. This is how far I got for general $k$.

Finishing the argument for the case $k = \mathbb{C}$: As we mentioned before, it is now sufficient to show that $h^{1,0} = 0$. The Hodge decomposition gives that $\dim_\mathbb{C} H^1(S, \mathbb{C}) = h^{1,0} + h^{0,1}$, so it is sufficient to show that $H^1(S,\mathbb{C}) = 0$ by the above. But all K3 surfaces are diffeomorphic and therefore it is sufficient to find a single K3 surface $S'$ with $H^1(S', \mathbb{C}) = 0$. From the adjunction formula and the cohomology of hypersurfaces in projective space, we see that any nice quartic surface in $\mathbb{P}^3$ is K3. But Lefschetz then tells us that $H^1$ vanishes for such a surface, since $H^1(\mathbb{P}^3,\mathbb{C})=0$.