Take all odd numbers that are non-divisible by 3. These are in the form of 2(3n)+1 and 2(3n+2)+1.
Simplify:
2(3n)+1
6n+1
Find 3n+1:
3(6n+1)+1
9n+2
Simplify:
2(3n+2)+1
6n+5
Find 3n+1:
3(6n+5)+1
9n+8
Take any number 1, 2, 4, 5, 7, or 8 mod 9. Repeatedly multiply by 2 until you get 2 or 8 mod 9. There are infinite number of numbers that fulfill this criteria, as mod multiplication is cyclic. Then find the n in 9n+8 or 9n+2 from this number. Substitute the n into 6n+1 or 6n+5. There are an infinite number of n and an infinite number of 6n+1 or 6n+5.
I don't really understand what you've written in the body of the question - you need to be clear about what, exactly, you're trying to prove at each stage - but it is indeed known that there are infinitely many odd numbers for which the Collatz conjecture holds. For example, suppose $2^n-1$ is divisible by $3$ (which will happen for infinitely many $n$ - if $2^n-1$ is not divisible by $3$, then $2^{n+1}-1$ is). Then $x={2^n-1\over 3}$ is odd (since $2^n-1$ is), and the Collatz sequence beginning with $x$ goes $$x, 2^n, 2^{n-1}, 2^{n-2}, . . . , 8, 4, 2, 1.$$