Let $K = \mathbb{Q}(\sqrt{7}, \sqrt{19})$, $L = \mathbb{Q}(\sqrt{7})$, and $M = \mathbb{Q}(\sqrt{19})$. Then we have that $3\mathcal{O}_L = \langle\sqrt{7} + 2 \rangle\langle2-\sqrt{7} \rangle$ and $3\mathcal{O}_M = \langle 4-\sqrt{19} \rangle\langle \sqrt{19} + 4\rangle$. Now, if $3\mathcal{O}_K = \mathfrak{P}_1^{e_1}\dots \mathfrak{P}_g^{e_g}$, since $G = \text{Gal}(K/\mathbb{Q}) \cong C_2 \times C_2$ and the decomposition groups $Z(\mathfrak{P_i})$ are conjugate to each other, it follows that $Z(\mathfrak{P}_i) = Z(\mathfrak{P_j})$ for each $i,j$. Therefore, we may denote $Z = Z(\mathfrak{P}_i)$. Let $\sigma, \tau \in G$ such that $\sigma: \sqrt{7} \mapsto - \sqrt{7}$ and $\tau : \sqrt{19} \mapsto - \sqrt{19}$, and observe that $\langle \sigma, \tau\rangle = G$. Because $3$ splits in $M$ and $L$, it must be that $2 \mid g = [G:Z]$ so that $Z \neq G$. Suppose that $g = 2$. Then the fixed field $K_Z$ for $Z$ must either be $M$, $L$, or $F = \mathbb{Q}(\sqrt{133})$. We also have that $3\mathcal{O}_F = \langle\sqrt{133}, 3 \rangle\langle \sqrt{133}-1 ,3\rangle$. Now, if we pick a $\mathfrak{P}_i$, looking at the ways $3$ can split in the degree $2$ sub-extensions implies that every element of $G$ moves one of the $\mathfrak{P}_i$. For example, if $\mathfrak{P_i} \cap K_Z = \langle 2 + \sqrt{7}\rangle$, then $\sigma(\mathfrak{P}_i)\cap K_Z = \langle 2 - \sqrt{7}\rangle$, and $\sigma(\mathfrak{P}_i) \neq \mathfrak{P}_i$. So none of the degree $2$ sub-extensions can be the fixed field for $Z$, and therefore $Z$ is trivial.
I think this is the right idea, but I'm having a little trouble explaining the part about every element of $G$ moving one of the $\mathfrak{P_i}$. Is there a better way to do this? I have a bad feeling that I'm overcomplicating this.
Thanks in advance!