If $T$ is bounded and $F$ has finite rank, what is the spectrum of $T+F$?

Short answer: $\sigma(T+F)$ is at most countable and so has empty interior.

Theorem (Spectral Theory of Linear Operators by V. Muller III.19.4): Let $T\in\mathcal{B}(X)$, let $G$ be a component of $\mathbb{C}\setminus\sigma_e(T)$. Then either $G\subset\sigma(T)$ or $G\cap\sigma(T)$ consists of at most countably many isolated points.

Combining the above with the essential spectrum, as you note, we get: $\sigma_e(T+F) = \sigma_e(T) \subset \sigma(T)$, which is finite. Then $G = \mathbb{C}\setminus\sigma_e(T+F)$ is a connected component and $\sigma(T+F)\setminus\sigma_e(T+F)$ is at most countable and all isolated.

You might also find that the spectrum must be finite but I can't think how to prove it.

It is always true if $T$ is self-adjoint. Here is a theorem that you might be interested:

If $T$ is self-adjoint, a complex number is in the spectrum of $T$ but not in its essential spectrum iff it is an isolated eigenvalue of $T$ of finite multipliticity.

The result can be found on page 32 of Analytic K-homology by Nigel Higson and John Roe.