On reducing the hardness of CNF-SAT to k-Clique
I don't know the answer to your specific question (it seems related to the question of whether W1=W[2]). 

But the algorithm you give in your question is subsumed by several other results. Using your definition of $N$, CNF-SAT is basically solvable in $O(1.1279^N)$ time, as in the paper by Wahlstrom (link goes to a google scholar page of papers that cite it). In particular, Wahlstrom shows that if $d$ is the average number of occurrences of a variable in the CNF, and no variable appears at most once, then CNF-SAT can be solved in $O(1.1279^{n(d - 2)})$ time. (There are other references with similarly strong running times.)

Update: Another reference is: 

Hirsch, E.A.: New worst-case upper bounds for SAT. J. Autom.
Reason. 24(4), 397–420 (2000)

This gives faster algorithms in terms of $m$ and your $N$ (which is typically called $\ell$). Namely, Hirsch gets $2^{0.30897m}$ and $2^{0.10299\ell}$ time.