On $\text{ETH}$ with $m$ as parameter: consequences of algorithm running in time $2^{\delta m}$ where $\delta \to 0$ as $k \to \infty$
This is an answer to the updated question (the original question seems harder).
Let $\mu'_k$ be the smallest constant such that $k$-SAT that has clauses of length exactly $k$ and no trivial clauses has a $O(2^{\mu'_k m})$ time algorithm.
Let $\mu_k$ be the smallest constant such that $k$-SAT with any clauses of length at most $k$ has an $O(2^{\mu_k m})$ time algorithm. I will prove that $\mu'_k = \mu_k$.
Proof: Clearly we have $\mu'_k \le \mu_k$. To prove that $\mu_k \le \mu'_k$ suppose we have an algorithm for $k$-SAT with nontrivial clauses of length exactly $k$ with time complexity $O(2^{\mu'_k m})$. To solve general $k$-SAT with $m$ clauses, we add $k$ extra variables to the input formula and $2^k-1$ extra clauses over these new variables to forbid all assignments except the all-false assignment. Now, these $k$ extra variables can be freely added to any clauses of length less than $k$ to make them length exactly $k$. Now we have a restricted instance with $m + 2^k-1$ clauses, and solve it in $O(2^{\mu'_k (m + 2^k-1)})$ time, which for constant $k$ is $O(2^{\mu'_k m})$ time.$\square$
It is easy to see that for $a \le b$ it holds that $\mu_a \le \mu_b$, and it is well-known (e.g. in section 14.1 of the book Parameterized Algorithms) that assuming ETH, $\mu_3 &gt; 0$. Therefore showing that $\mu'_k \rightarrow 0$ as $k \rightarrow \infty$ would break ETH.