Operator that commutes with projections

Maybe a quicker way to see this is, if $TP_m = P_mT$ for all $m \neq n$ then $TP = PT$ where $P =\sum_{m\neq n} P_m = I - P_n$. Since $T$ commutes with $I$, it must therefore commute with $P_n$.

The answer is no. Indeed, without loss of generality (wlog), $n=0$, so that $TP_m=P_mT$ for all $m=1,2,\dots$ (assuming here that $\mathbb{N}_0=\{0,1,\dots\}$). Since $P_0=I-\sum_{m\ge1}P_m$ and $T I=IT$, it follows that $TP_0=P_0T$. (Here, of course, $I$ is the identity operator and the operator $S:=\sum_{m\ge1}P_m$ is defined by the condition that $Sx:=\sum_{m\ge1}P_m x$ for all $x\in\ell^2(\mathbb{N}_0)$.)