Resolution on 3-SAT instance yields in polynomial many resolvents A SAT instance in CNF with $n$ variables has at most $2^n$ resolvents, therefore the resolution method is not in polynomial time.

Considering a 3-SAT instance, we have at most $n^3 + n^2 + n$ many resolvents. Therefore, we can decide 3-SAT in polynomial time by resolution.

I am sure, my second statement is wrong, because otherwise we would have $P = NP$. But I can not see my mistake. Can someone bring clarification here?
There are only that many clauses but that only gives an upper bound for the time to test a single candidate solution.  There are still $2^n$ candidates to test.
Using some algorithms like DPLL or CDCL SAT solvers leads to much less than $2^n$ candidate testing but still an exponential function.

Also, in local search algorithms best bounds are about $1.333^n$ by Schöning's 1999 paper.