Let $wt(n)$ be [A000120][1], number of $1$'s in binary expansion of $n$ (or the binary weight of $n$)
and
$$n=2^{t_1}(1+2^{t_2+1}(1+\dots(1+2^{t_{wt(n)}+1}))\dots)$$
Then we have an integer sequence given by
$$a(n)=\sum\limits_{j=0}^{2^{wt(n)}-1}m^{wt(n)-wt(j)}\prod\limits_{k=0}^{wt(n)-1}(1+wt(\left\lfloor\frac{j}{2^k}\right\rfloor))^{t_{k+1}+1}, a(0)=1$$
Let
$$s(n,m)=\sum\limits_{k=0}^{2^n-1}a(k)$$
then I conjecture that for any $m$
$$s(n,m)=\sum\limits_{k=0}^{n+1}k!{n+1\brace k}(m+1)^{n-k+1}$$
Is there a way to prove it?

Similar questions:

 - [Recurrence for the sum][2]
 - [Pair of recurrence relations with $a(2n+1)=a(2f(n))$][3]
 - [Sequence that sums up to INVERTi transform applied to the ordered Bell numbers][4]
 - [Sequences that sums up to second differences of Bell and Catalan numbers][5]


  [1]: https://oeis.org/A000120
  [2]: https://mathoverflow.net/questions/405174/recurrence-for-the-sum
  [3]: https://mathoverflow.net/questions/406902/pair-of-recurrence-relations-with-a2n1-a2fn
  [4]: https://mathoverflow.net/questions/407290/sequence-that-sums-up-to-inverti-transform-applied-to-the-ordered-bell-numbers
  [5]: https://mathoverflow.net/questions/407758/sequences-that-sums-up-to-second-differences-of-bell-and-catalan-numbers