number theory - Euler's totient function determining $phi(m)$= 2 using the product formula.

Solving $\phi(m)$ =2 , to find that the only possible types of m are 3,4, and 6

I have considered the product and found $p_i$ = 2 or 3 and $a_i$ = 2 or 1 and this I have found using cases 1|2 and 2|2.

How can I use my results combining the two cases that the only possible such m are if 3,4, or 6.

edit: I am considering the case of when m and n are coprime and I know the possible solutions are (3,4) and (4,3) . For self, I am wondering about how I find these m after obtaining my cases 1|2 and 2|2.