Let $X$ be a Banach space and $\Omega\subseteq\mathbb{C}$ be open, and let $T$ be as in the title with $\sigma(T)\subset\Omega$ and $f\in H(\Omega)$.
We can define $f_{n}(\lambda)=\lambda^{n}$. Then $\Phi_{T}(f_{n})=T^{n}$. Thus we have that $\Phi_{T}(f_{2})=T^{2}=0$. Now, let $x\in X$, with $x\ne 0$, then $\Phi_{T}(f)x=f(\alpha)x$, for $\alpha\in\mathbb{C}$; which implies that every eigenvector of $T$, with eigenvalue $\alpha$, is also an eigenvector of $\Phi_{T}(f)$, with eigenvalue $f(\alpha)$.
Note that $\Phi_{T}(f):=f(T)$.
I guess I would want to use the spectral mapping theorem from here, or am I even going about this in the right way? I feel that I've maybe overcomplicated things a tad.