I don't know about the approach in its entirety, but I can confirm the congruence.
If 2^{n1} == 1 (mod n), and p is a prime for which pn, then 2^{n1} == 1 (mod p).
Consequently, if m is the least positive integer for which 2^{m} == 1 (mod p) [m is the multiplicative order of 2 (mod p)] then m divides n1.
We thus have n == 1 (mod m). Since 2^{p1} == 1 (mod p) m also divides p1, so p == 1 (mod m), and n == p (== 1) (mod m).
Since by hypothesis p divides n, we have n == p (== 0) (mod p).
Now m divides p1, so m and p are relatively prime. By CRT we have n == p (mod mp).
