This is a method I have been playing with. It has $T^x$ behavior, but this may not be as bad as you think.
My solver works for monotone, linear Exactly 1 in 3 SAT, ml-X3SAT. Monotone means all variables are positive and linear means no two clauses have more than one variable in common. ml-X3SAT is NP-Complete. I also assume we can apply the Two Singleton reduction rule. A singleton is a variable that appears exactly once in the instance. We can remove any clause with two singletons, like $(a, s_1, s_2)$, and assume $s1=NOT(a), s_2=False$. This rule allows us to assume no clause has more than one singleton. If no variable occurs in more than two clauses then ml-X3SAT reduces to a Perfect Matching instance solvable in polynomial time.
Choose a variable that appears in $k\geq 3$ clauses. I call these $k$ clauses an index. $k$ clauses that share a variable have $2k+1$ variables and $2^k+1$ satisfying assignments. It turns out the worst case is $k=3$ so assume the first index has three clauses. Remove all clauses that share a variable with one of the three clauses in the first index. Now choose a variable in the remaining instance that appears in three or more clauses. Make clauses with this variable the second index. Continue this process of selecting indices until no variable appears in more than two of the remaining clauses.
By construction, no index shares a variable with any other index. If we choose one satisfying assignment from each index then we will either find a contradition or the remaining clauses reduce to a Perfect Matching instance. Because no clause has more than one singleton, any satisfying assignment of the $k$ clauses in an index will also reduce another $k$ clauses. Each clause in an index must share a variable with a non-index clause. If $m$ is the number of clauses in the original instance then we will not need more than $m/2k$ indices.
If we assume $k=3$ for every index then we can solve an instance in $O(9^{m/6})$ or $O(2^{.53m})$. This isn't nearly as good as the best known, but it is not bad for a simple analysis. It shows you want to choose a small number of indices where each index has a large number of clauses and variables and a small number of satisfying assignments. The number of satisfying assignments, $T$, is not as important as the number of indices, $x$, in $T^x$. Any method of reducing the number of indices is good. For example, after choosing an index we can apply reduction rules to the remaining clauses to minimize the remaining instance.
Answer to question:
$m$ is the number of clauses in the entire $ml-X3SAT$ instance. $k$ is the number of clauses in each index. We assume all of these clauses in an index share a variable in common. Because we assume the instance is "linear", these clauses can't share more than one variable. We also assume $k \geq 3$ because if no variable is in more than two clauses then the instance reduces to perfect matching and is solvable in polynomial time.
$k$ clauses with a variable in common have $2^k+1$ satisfying assignments. There is one solution where the shared variable is true and all of the other variables are false. All of the other solutions have the shared variable false and one of the other two variables in each clause is true. This gives us $2^k$ more solutions.
Because no clause has more than one singleton, we know each clause in the index shares a variable with at least one non-index clause. Each satisfying assignment of the index clauses removes at least $2k$ clauses. This means we need no more than $m/2k$ indexes to "cover" the entire instance. We will need to examine no more than $(2^k+1)^{m/2k}$ assignments to find any solution.
When we convert this to base 2 the exponent will have $log(2^k+1) \approx k$ giving us $2^{mk/2k} = 2^{m/2}$ for large $k$. $k=3$ is the worst case because the "$1$" in $2^k+1$ is large compared to $2^3$. For example for $k=10$ we get $1025^{m/20} = 2^{.50007m}$.
