This is in principle correct (although it's not $O(n!)$ but essentially just $O(2^n)$), because all of the operations involved are linear. For convenience, I'll use $\frac12(3n+1)$ as the 'growth' operator rather than the usual $3n+1$ (the two are equivalent since every $3n+1$-step is always followed by a halving); suppose we want to know whether there's a cycle that starts on an odd number and goes growth-half-growth. Then we solve the equation $n = g(h(g(n)))$, where $g(n)=\frac12(3n+1)$ and $h(n)=\frac n2$; this is equivalent to $n=\frac12\left(3(\frac14(3n+1))+1\right)$, or $n=\frac98n+\frac78$, or $n=-7$; thus, we know that no cycle can follow this pattern. More broadly, whatever the pattern we choose, we can 'translate' that pattern down to a single linear function, and so the equation $n=P(n)$ (where $P(n)$ is the function corresponding to the given pattern) can be solved for its (unique) solution, which can then be tested to see whether it's a positive integer (and whether it satisfies the constraints - that is, whether the sequence of Collatz operations applied to the number matches the sequence that was hypothesized).
But there are two catches: first, a potential cycle could be arbitrarily long, so this doesn't actually reduce the problem to a finite computation - it just allows for a different computation. Secondly, it's possible for there to be no other cycles but for the conjecture to still be false - in particular, it could be that there's only one cycle but that there's a sequence that grows without bound, and your method would never detect this case.
That said, this is an angle on the conjecture that's been investigated before - for instance, the paper "The 3x+1 problem and its Generalizations" from Jeffrey Lagarias talks about some related ideas and I certainly encourage you to study more deeply in those directions.
