I knew this problem was not out of the blue! Of course one can put $m$ instead of $2^n$. It's easier when $m$ is a power of $2$. https://dgrozev.wordpress.com/2020/07/22/a-family-of-sets-bulgarian-tst-for-imo-2020-p3/ Yeah, it was given for $m=100$, but the same argument can be used for any $m$ - see the recurrence formula at the end.

Solution 1. The required minimum is $n\cdot 2^n +1$. Collections of this size, satisfying the conditions in the statement, are recursively obtained as follows: The collection $\mathcal{C}_1$ consisting of $\emptyset, \{1\}, \{2\}$ clearly fits the bill for $n = 1$. For a larger $n{}$, consider a collection $\mathcal{C}_{n-1}$ fitting the bill for $n-1$, and form the collection $\mathcal{C}_{n-1}'$ of size $|\mathcal{C}_{n-1}|$, consisting of all sets of the form \[C\cup\{2^{n-1}+1,2^{n-1}+2, \ldots , 2^n\}\]for $C\in\mathcal{C}_{n-1}$. Next, collect all subsets of the form $\{2^{n} + 1-c: c\in C\}$ for $C\in\mathcal{C}_{n-1}'$ to form another collection $\mathcal{C}_{n-1}''$ of size $|\mathcal{C}_{n-1}|$ disjoint from $\mathcal{C}_{n-1}'$ Finally, form $\mathcal{C}_n$ by collecting together the members of $\mathcal{C}_{n-1}'\cup\mathcal{C}_{n-1}''$ along with the empty set and the $2n-2$ sets \begin{align*}\{1\}\qquad &\phantom \qquad \{2^{n-1}+1\} \\ \{1,2\}\qquad &\phantom \qquad \{2^{n-1}+1,2^{n-1}+2\} \\ \vdots\qquad &\phantom \qquad\vdots \\ \{1,2,\ldots,2^{n-1}-1\}\qquad &\phantom \qquad \{2^{n-1}+1,2^{n-1}+2,\ldots,2^n-1\}\end{align*}none of which lies in $\mathcal{C}_{n-1}'\cup\mathcal{C}_{n-1}''$. It is now readily checked that $\mathcal{C}_n$ fits the bill for $n{}$. Write $X = \{1, 2, \ldots , 2^n\}$, and let $\mathcal{C}$ be a collection of subsets of $X{}$ satisfying the conditions in the statement. We now show that $|\mathcal{C}|\geqslant |X| \cdot \log_2 |X| + 1$, whence the conclusion. Clearly, $\mathcal{C}$ contains the empty set, and we may and will assume that $X{}$ is not a member of $\mathcal{C}$. For every non-empty $C{}$ in $\mathcal{C}$, choose once and for all an element $x_C$ of $C{}$ such that $C\setminus\{x_C\}$ is a member of $\mathcal{C}$. Remove all superfluous members of $\mathcal{C}$, i. e., all $\subseteq$-maximal members of size less than $|X|-1$. Repeat until the resulting collection has no more such. The outcome is a collection, also denoted $\mathcal{C}$, whose maximal members are all of size $|X|-1$; this $\mathcal{C}$ also satisfies the conditions in the statement. The parthenogenetic (rooted) tree on vertex set $\mathcal{C}$ is defined by letting every non-empty $C{}$ in $\mathcal{C}$ have parent $C\setminus\{x_C\}$. Thus, genesis starts at the empty set to end at all $(|X|-1)$-element subsets of $X{}$, the leaves of the tree. For every vertex $C$, let $h_C$ be the vertex distance of $C$ to the nearest leaf, let $s_C$ be the number of leaves in the subtree rooted at $C$, and let $v_C$ be the size of the vertex set of this subtree. Lemma. If $C$ is a member of $\mathcal{C}$, then $h_C\geqslant s_C$, and $v_C\geqslant s_C\log_2 s_C+h_C-s_C+1$. Consequently, \[|\mathcal{C}|=v_\emptyset\geqslant s_\emptyset\log_2 s_\emptyset+h_\emptyset-s_\emptyset+1=|X| \cdot \log_2 |X| + 1.\]Proof. To prove the first inequality, notice that $h_C = |X|-|C|$ which is precisely the total number of subsets of $X{}$ of size $|X|-1$ containing $C{}$. The latter form a collection of leaves containing the leaves of the subtree rooted at $C$, so $s_C \leqslant |X| - |C| = h_C$. To prove the second inequality, induct on $h_C$. The base case corresponds to $C{}$ being a leaf: $h_C = 1, s_C = 1$, and the inequality is clear. For the induction step, let $C_1, \ldots, C_m$ be the sons of $C$, so $h_{C_i} = h_{C}-1$ for $i = 1, \ldots , m$, $s_{C_1}+\cdots+s_{C_m}=s_C$ and $v_{C_1}+\cdots v_{C_m}=v_C-1$. Refer to the induction hypothesis to write \begin{align*}v_C&=1+\sum_{i=1}^mv_{C_i}\geqslant 1+\sum_{i=1}^m\left(s_{C_i}\log_2 s_{C_i}+h_{C_i}-s_{C_i}+1\right) \\ &=\sum_{i=1}^ms_{C_i}\log_2 s_{C_i}+\sum_{i=1}^mh_{C_i}-\sum_{i=1}^ms_{C_i}+m+1=\sum_{i=1}^ms_{C_i}\log_2 s_{C_i}+\sum_{i=1}^m(h_C-1)-s_C+m+1 \\ &=\sum_{i=1}^ms_{C_i}\log_2 s_{C_i}+mh_C-s_C+1 =\left(\sum_{i=1}^ms_{C_i}\log_2 s_{C_i}+(m-1)h_C\right)+h_C-s_C+1 \\ &\geqslant \left(\sum_{i=1}^ms_{C_i}\log_2 s_{C_i}+(m-1)s_C\right)+h_C-s_C+1.\end{align*}It is sufficient to show that the term in large parentheses is at least $s_C \log_2 s_C$. To this end, notice that the function $x\mapsto x \log_2 x$ for $x > 0$, is convex, to write \[\sum_{i=1}^ms_{C_i}\log_2 s_{C_i}\geqslant\left(\sum_{i=1}^ms_{C_i}\right)\log_2\left(\frac{1}{m}\sum_{i=1}^ms_{C_i}\right)=s_C\log_2\frac{s_C}{m}.\]Finally, notice that $\log_2(x/m) + m - 1 = \log_2 x - \log_2 m + m - 1 \geqslant \log_2 x$, where $x > 0$ and $m$ a positive integer, since $2^{m-1}\geqslant m$, to conclude that \[\sum_{i=1}^ms_{C_i}\log_2 s_{C_i}+(m-1)s_C\geqslant s_C\log_2\frac{S_C}{m}+(m-1)S_C\geqslant s_C\log_2 s_C.\]Solution 2. We solve a more general problem, where $2^n$ is replaced by an arbitrary positive integer $k{}$. Say that a collection $\mathcal{C}$ of subsets of a $k{}$-element set $X{}$ is $k{}$-good if it satisfies the requirements of the problem, that is, $\mathcal{C}$ contains all $(k-1)$-element subsets of $X{}$, and every non-empty set $C\in\mathcal{C}$ contains an element $c{}$ such that $C\setminus\{c\}$ also lies in $\mathcal{C}$. We are interested in the smallest size of a $k{}$-good collection. We will show that for $k = 2^d + s$, where $0\leqslant s \leqslant 2^d$, the answer is $f(k) = 1 + (d + 2)s + d \cdot 2^d$. Notice here that two representations of a power of $2{}$ yield the same value of $f(k)$. Also, it is easy to see that the function $f{}$ satisfies the recurrence relation \[f(k)=f(\lfloor k/2\rfloor)+f(\lceil k/2\rceil)+(k-1).\]The recurrence relation allows a recursive construction of a $k{}$-good collection consisting of $f(k)$ sets along the lines in the previous solution. We now show that such a collection cannot contain less than $f(k)$ sets, by induction on $k{}$. The base case $k = 1$ is trivial. Consider a $k{}$-good family $\mathcal{C}$ consisting of $N{}$ sets, where $k = 2^d + s$ with $1 \leqslant s \leqslant 2^d$ (by an above remark, we may exclude the value $s = 0$). For any non-empty set $C\in\mathcal{C}$, choose some $c\in C$ such that $C\setminus\{c\}\in\mathcal{C}$, and define $g(C) = c$. The domain of $g{}$ contains $N-1$ sets, while the range contains at most $k{}$ elements; hence some $c\in X$ is assumed at least $p=\lceil(N-1)/k\rceil$. Fix such an element $c\in X$. Let now $\mathcal{C}' =\{C\setminus\{c\}: C\in\mathcal{C}\}$. For any $C\in\mathcal{C}$ with $g(C) = c$, the sets $C{}$ and $C\setminus\{c\}$ provide the same set in $\mathcal{C}'$. Hence $|\mathcal{C}'|\leqslant N-p$. Clearly, $\mathcal{C}'$ is a $(k-1)$-good collection of subsets in $X' = X\setminus\{c\}$; moreover, $\mathcal{C}$ contains $X'{}$ itself which can be removed harmlessly, to obtain a good collection $\mathcal{C}''$ consisting of $N-1-p=\lfloor(N-1)(k-1)/k\rfloor$ sets. Thus \[\left\lfloor(N-1)\frac{k-1}{k}\right\rfloor\geqslant f(k-1);\]alternatively, but equivalently, $(k-1)(N-1)\geqslant kf(k-1)$. Finally, it suffices to check that this inequality fails for $N = f(k)-1 = d \cdot 2d + (d + 2)s$. In other words, \[(2^d+s-1)(d2^d+(d+2)s-1)<(2^d+s)(d2^d+(d+2)s-1-d);\]alternatively, but equivalently, $-(d2^d+(d+2)s-1)<-d(2^d+s)$ which is the case if and only if $1 < 2s$, and this is clearly true. This completes the induction step and concludes the proof.

Very nice problem! So nice, in fact, that it appeared on three separate contests The answer is $n2^n+1$. By taking the complement of everything, we wish to find the smallest collection of subsets $\mathcal{C}$ containing every $1$-element subset such that for every element $[2^n] \neq S \in \mathcal{C}$ there exists some $c \not \in S$ such that $S \cup \{c\} \in \mathcal{C}$. We use a recursive construction. For $n=0$ there is nothing to do. Now, to construct for $n>0$, we take two copies of the construction for $n-1$ on $\{1,\ldots,2^{n-1}\}$ and $\{2^{n-1}+1,\ldots,2^n\}$. Then for $1 \leq i<2^{n-1}$ we add the sets $\{1,\ldots,2^{n-1}+i\}$ and $\{2^{n-1}+1-i,\ldots,2^n\}$, and finally we add $\{1,\ldots,2^n\}$. This clearly works, and uses a total of $$2((n-1)2^{n-1}+1)+2(2^{n-1}-1)+1=n2^n-2^n+2+2^n-2+1=n2^n+1$$sets. Note that To prove that at least $n2^n+1$ sets are needed, I will show that for any $0 \leq i<n$ we need at least $2^n$ sets whose sizes are in $[2^i,2^{i+1})$. Note that the conditions imply that for any $1 \leq k \leq 2^n$, every element of $[2^n]$ is in some element of $\mathcal{C}$ with size $k$. Thus this claim finishes the problem, since $[2^n] \in \mathcal{C}$ is forced. Fix an arbitrary $i$. Construct a graph on the elements of $\mathcal{C}$ with size in $[2^i,2^{i+1})$, drawing a directed edge $x \to y$ if $x=y \cup \{c\}$ for some $c \not \in y$. Split the graph into "layers" based on size, with the lowest layer containing the smallest sets; thus edges go from a layer to the one below it. The conditions imply that every vertex not on the topmost layer has indegree at least $1$. Now by deleting some edges, transform the graph into a union of disjoint trees rooted at a topmost vertex. If some vertex not on the bottom layer now has outdegree $0$, it can be deleted from $\mathcal{C}$ and thus the graph; repeatedly doing this leaves us with trees where every "branch" extends from top to bottom. Note that this means the number of vertices in each layer is nondecreasing as we move downwards. Observe that the union of all sets in $\mathcal{C}$ with size $2^i$ should be $[2^n]$. Fix an arbitrary tree, and consider its lowest vertex that's "alone" in its layer (i.e. this layer only contains one vertex—itself). By definition, the union of all sets of size $2^i$ in a given tree (i.e. the bottom vertices) should have at most as many elements as its corresponding set. It is clear that if this lowest vertex has size $k$ then the tree has at least $2^i+(k-2^i)=k$ vertices. Thus by summing over all trees, we need at least $2^n$ vertices in total, which finishes the problem. $\blacksquare$