wikipedia says: there can be also cases where is still possible to communicate through the quantum channel encoding more than the classical information
What I propose is to switch between deterministic and indeterministic process, the former encoding 0, the latter encoding 1. Maybe indeterministic process counts as more than the classical information. Also since it's indeterministic, it has a small chance of miscommunication. Since communication isn't precisely certain, maybe the theorem treats it as failure.
Wikipedia is referring to superdense coding, which is not superluminal and still needs you to physically send a particle from place to place to transfer the information.
"an entangled state (e.g. a Bell state) is prepared using a Bell circuit or gate by Charlie, a third person. Charlie then sends one of these qubits (in the Bell state) to Alice and the other to Bob. Once Alice obtains her qubit in the entangled state, she applies a certain quantum gate to her qubit depending on which two-bit message (00, 01, 10 or 11) she wants to send to Bob. Her entangled qubit is then sent to Bob who, after applying the appropriate quantum gate and making a measurement, can retrieve the classical two-bit message." (emphasis mine)
Alice physically transmits her qubit to Bob, so it's not superluminal and doesn't break the no-communication theorem.
Okay, I found my error: deterministic measurement I proposed is impossible, it works only for factorized state, but entangled state produces second spooky result with uniform distribution, so collapsed photon is indistinguishable from noncollapsed.
What I propose is to switch between deterministic and indeterministic process, the former encoding 0, the latter encoding 1. Maybe indeterministic process counts as more than the classical information. Also since it's indeterministic, it has a small chance of miscommunication. Since communication isn't precisely certain, maybe the theorem treats it as failure.