In usual complex analysis, a pole $z_0\in \mathbb{C}$ of a function $f$ that is holomorphic on a punctured disk $0<|z-z_o|<R$ is defined as a zero of function $\frac{1}{f}$. But in spectral theory, where we consider the singularities of the resolvent $f: z\mapsto (z-A)^{-1}$, where $A\in \mathcal{B}(X)$ is a bounded operator. Here $f$ is defined on $\rho(A)$, which is an open set, and is holomorphic.
How should we define a point $z_0\in \sigma(A)$ to be a pole in this context??
Here, $\frac{1}{f}$ seems to be analogous to $z\mapsto z-A$, but $z_0-A$ is never the zero map unless $A=z_0$. So, here, should I just understand this term pole as in "if there exists $n\in\mathbb{N}$ such that $(z-z_0)^n (z-A)^{-1}$ is holomorphic on the entire disk $|z-z_0|<R$, and if $z_0$ remains a singularity for $(z-z_0)^{n-1} (z-A)^{-1}$"?
First of all: A pole of the resolvent mapping is never an element of the resolvent set $\rho(A)$, since the resolvent mapping (the map $z \mapsto (z -A)^{-1}$) is holomorphic on $\rho(A)$.
One way to define a pole is as follows (this is also often done in "regular" complex analysis): Suppose $\lambda$ is an isolated point of the spectrum $\sigma(A)$ of $A$. Then $\lambda$ is a pole of order $n$ of the resolvent map, if there exist operators $B_1, \dots, B_n \in \mathcal{B}(X)$ with $B_n\neq 0$ so that the function $z \mapsto (z-A)^{-1} - \sum_{k=1}^n B_k (z-\lambda)^{-k}$ has a removable singularity at $\lambda$.
This is the same as requiring, that the Laurent expansion of the resolvent around $\lambda$ has only finitely many terms with negative power and that the lowest power term (with nonzero coefficient) has power $-n$.
The above definition works more generally for any Banach space valued holomorphic function.