Let's say I have $y \in \mathbb{R} \rightarrow x \leftarrow z \in \mathbb{R}^k $. I think I model $p(x,y,z) = p(x|y,z)p(z)p(y)$?
From that, I think that bayes rule says that
$$ p(z,y|x) = \frac{p(x|y,z)p(z)p(y)}{p(x)} $$
Can I also derive that $$ p(z,y|x) = \frac{p(x|y,z)p(z)p(y|x)p(x)}{p(x)} = p(x|y,z)p(z)p(y|x) \text{?} $$
From your derivation, you are implicitly implying that: $$p(y)=p(y|x)p(x)$$ which is true unless $p(x,y)=p(y)$, because $p(x,y)=p(y|x)p(x)$
So do you also want to assume $p(x,y)=p(y)$? It is up to you, as you already have some assumption about joint distribution of $x,y,z$