I used Magma to point search on $C/K$ up to a height of $1000$ and it appears that $C(K) = \emptyset$. If that's true, then one can probably use the Mordell-Weil sieve to prove it. Here's a bit more detail.
The curve $C$ has four automorphisms defined over $\mathbb{Q}$ and for one of these (the map $(x,y) \mapsto (1/x,y/x^{3})$), the quotient curve has genus $1$. In particular, there is a map from $\phi : C \to E$, where $E : y^{2} = x^{3} - 651x - 12742$. This curve $E$ has rank $1$ over $K$.
For a finite set $S$ of prime ideals of $\mathcal{O}_{K}$ (all of which are primes of good reduction for $E/K$), one can write down the commutative diagram
$$
\require{AMScd}
\begin{CD}
C(K) @>>> E(K)\\
@VVV @VV{\beta}V\\
\prod_{\mathfrak{p} \in S} C(\mathbb{F}_{\mathfrak{p}}) @>\alpha>> \prod_{\mathfrak{p} \in S} E(\mathbb{F}_{\mathfrak{p}})\\
\end{CD}
$$
The horizontal maps in this diagram use the map from $C$ to $E$, while the vertical maps use reduction modulo the prime ideals in $S$.
If $P \in C(K)$ is a point, then the image of $\alpha$ and the image of $\beta$ (as subsets of $\prod_{\mathfrak{p} \in S} E(\mathbb{F}_{\mathfrak{p}})$) have a non-trivial intersection. So if one can find a set $S$ of primes for which the image of $\alpha$ and the image of $\beta$ are disjoint, this proves that $C(K)$ is empty.
For more about this, I recommend The Mordell-Weil sieve: Proving non-existence of rational points on curves by Nils Bruin and Michael Stoll.
