No another leap year would be fine. But yeah a very common (wrong) pattern is, if you want to find the same day a year in advance, is to do `d.replace(year=d.year + 1)`, and that would break on Feb 29 only, so one day every 4 years. It's a very common pattern unfortunately.
It returns a new datetime object which includes its own validity check and raises an exception for a bad leap day the same way it would for March 32nd or Blurnsuary 12th. (However, year must be an integer in [1, 9999].)