The standard solution is to have a bounded utility function, and it seems like it fully solves this reformulation as well. There may also be some other solutions that work for this, but I’m sufficiently unsure about all the notation that I’m not very confident in them.
You’re right. Somehow I completely missed Pascal’s Mugging for bounded utility functions
EDIT: I obviously didn’t miss it, because I commented there. What I did do was not understand it, which is quite a bit worse.