^                     # N = tail = input
(?=(x(x*?))(\1\1)+$)  # \1 = N / {largest odd divisor of N}
                      #    == {largest power of 2 divisor of N}; \2 = \1 - 1
((x*)(?=\5$))+        # tail = N / {largest power of 2 divisor of N}
                      #      == {largest odd divisor of N}
(?!(xx+)\6+$)         # Assert tail is prime
\1\2$                 # Assert tail == \1*2 - 1; if this fails to match, it will
                      # backtrack into the "((x*)(?=\5$))+" loop (effectively
                      # multiplying by 2 repeatedly), but this will always fail
                      # to match because every subsequent match attempt will be
                      # an even number, and "\1\2$" can only be odd.