Erdős problem 403

Does the equation $$2^m=a_1!+\cdots+a_k!$$ with $a_1<a_2<\cdots <a_k$ have only finitely many solutions?

Asked by Burr and Erdős. Frankl and Lin [Li76] independently showed that the answer is yes, and the largest solution is $$2^7=2!+3!+5!.$$ In fact Lin showed that the largest power of $2$ which can divide a sum of distinct factorials containing $2$ is $2^{254}$, and that there are only 5 solutions to $3^m=a_1!+\cdots+a_k!$ (when $m=0,1,2,3,6$).

See also [404].

A solution is encoded below as a pair $(m, s)$ where $s$ is the finite set $\{a_1 < a_2 < \cdots < a_k\}$ of positive integers, so the distinctness of the $a_i$ is given by set membership. The empty set contributes no solutions since $2^m \geq 1 > 0$.

The linked proof gives more than finiteness: it classifies the solutions outright, as $(0,\{1\})$, $(1,\{2\})$, $(3,\{2,3\})$, $(5,\{2,3,4\})$ and $(7,\{2,3,5\})$, so the set below has exactly five elements.