Let $\mathcal{M}(H^\infty(\mathbb{D}))$ denote the spectrum of the Banach algebra $H^\infty$ and $\mathcal{M}_z(H^\infty(\mathbb{D}))$ the fiber over $z\in \mathbb{D}$, i.e. $\{\varphi\in \mathcal{M}:\varphi(\text{Id})=z\}$. It is well known that for any $z\in \partial \mathbb{D}$, $\mathcal{M}_z$ contains an embedded copy of $\beta \mathbb{N}$ so it is not metrizable and it contains a non-separable subset. This suggests that $\mathcal{M}_z$ is itself not separable, but I have not been able to prove it. So my question is:

Is $\mathcal{M}_z$ separable?

Let $\mathcal{M}(H^\infty(\mathbb{D}))$ denote the spectrum of the Banach algebra $H^\infty$ and $\mathcal{M}_z(H^\infty(\mathbb{D}))$ the fiber over $z\in \mathbb{D}$, i.e. $\{\varphi\in \mathcal{M}:\varphi(\text{Id})=z\}$. It is well known that for any $z\in \partial \mathbb{D}$, $\mathcal{M}_z$ contains an embedded copy of $\beta \mathbb{N}-\mathbb{N}$ so it is not metrizable and it contains a non-separable subset. This suggests that $\mathcal{M}_z$ is itself not separable, but I have not been able to prove it. So my question is:

Is $\mathcal{M}_z$ separable?