I would like to show, that if I have a linear operator $T\in\mathcal{B}(H)$, and a compact operator $K\in\mathcal{C}(H)$ for which $r(K)>0$ ($r$ is the spectral radius - i.e. $K$ is non-quasinilpotent), and $TK=KT$, then $\sigma_p(T)\neq\emptyset$ ($\sigma_p(T)$ is the point spectrum of $T$).
I don't have any relevant results yet, but in my opinion, the following theorem will be useful:
Theorem: If $K\in\mathcal{C}(H)$, and $\lambda\in\sigma_p(K)\backslash\{0\}$, then $$\dim ker(K-\lambda I)=\dim ran(K^*-\overline{\lambda}I).$$
Because $K$ is compact, and $\sigma(K)\ne \{0\}$, then there exists $\lambda \ne 0$ such that $\mathcal{N}(K-\lambda I)\ne \{0\}$. This null space is finite-dimensional because $K$ is compact and $K=\lambda I$ on $\mathcal{N}(K-\lambda I)$. Furthermore, because $KT=TK$, one has $(K-\lambda I)x=0\implies (K-\lambda I)Tx=0$. Hence, $$ T : \mathcal{N}(K-\lambda I)\rightarrow \mathcal{N}(K-\lambda I). $$ Because you're working on a complex space, the restriction of $T$ to $\mathcal{N}(K-\lambda I)$ must have an eigenvalue. So there exists $\mu\in\mathbb{C}$ and $x\in\mathcal{N}(K-\lambda I)$ such that $x\ne 0$ and $Tx=\mu x$. Hence $\sigma_{p}(T)\ne \emptyset$.