Tuples of sets $A_1, ... , A_k \in \mathcal{P}([n])$ with $\bigcup_{i=1}^{k}{A_i}=[n]$ and $A_i \neq A_j$, $1 \le i \lt j \le k$

Another, non direct, way to answer the linked question is by using binomial inversion for the numerical part and using an inclusion-exclusion argument, so that if $$\binom{2^m}{2}=\sum_{n=0}^m {m \choose n}\frac{3^n-1}{2},$$ then $$\frac{3^m-1}{2}=(-1)^m\sum _{n=0}^m\binom{m}{n}(-1)^n\binom{2^n}{2}.$$ In this way, I think you are looking for $$(-1)^m\sum _{n=0}^m\binom{m}{n}(-1)^n\binom{2^n}{k}.$$ which agrees with your computation. You can check this combinatorially using inclusion-exclusion by considering which element is missing in your decomposition.

Edit: Consider then $$(-1)^m\sum _{n=0}^m\binom{m}{n}(-1)^n\binom{2^n}{k}=\binom{2^{m}}{k}-\sum _{n=1}^{m}\binom{m}{n}(-1)^{n-1}\binom{2^{m-n}}{k},$$ this corresponds to doing $$\left |\mathcal{A}\setminus \bigcup _{\ell=1}^m\mathcal{A}_{\ell}\right |,$$ where $\mathcal{A}$ is all possible ways to get $k$ different subsets $\{A_i\}$ of $[n]$ and $\mathcal{A}_{\ell}$ is the number of ways to get $k$ different subsets of $[n]$ such that $\ell$ is in non of them.

Show that $|\mathcal{A}|=\binom{2^m}{k},$ and that if $X\subseteq [m],$ then $$\left |\bigcap _{x\in X}\mathcal{A}_x\right |=\binom{2^{n-|X|}}{k}.$$ Notice that $$\mathcal{A}\setminus \bigcup _{\ell=1}^m\mathcal{A}_{\ell}$$ is exactly the objects you want. You eliminate every time their union was not exactly $[n].$