Recursion for the A266328 by analogy with non-standard recursion for factorials

Question

• Let $a(n)$ be A266328 i.e. an integer sequence with exponential generating function $$ A(x)=\exp\int B(x) \,dx $$ such that $$ B(x)=\exp(-x)\exp\int A(x) \,dx $$ where the constant of integration is zero.
• Let $$ R(n,q,m)=R(n-1,q+1,m)+\sum\limits_{j=0}^{q-m} \binom{q+1}{j}R(n-1,j,m), \\ R(0,q,m)=1. $$

I conjecture that $$ R(n,0,0)=(n+1)!, \\ R(n,0,1)=a(n+1). $$

Here is the PARI/GP prog to check it numerically:

a(n) = my(A=1+x, B=1+x); for(i=0, n, A = exp( intformal( B + x*O(x^n) ) ); B = exp( intformal( A - 1 ) ) ); n!*polcoeff(A, n)
R_upto(n,m)=my(v1, v2, v3, v4); v1=vector(n+1, i, 1); v2=v1; v3=vector(n+1, i, 0); v3[1]=1; v4=vector(n, i, vector(i+1, j, binomial(i, j-1))); for(i=1, n, for(q=0, (n-i), v2[q+1]=v1[q+2]+sum(j=0, q-m, v4[q+1][j+1]*v1[j+1])); v1=v2; v3[i+1]=v1[1]); v3
test1(n)=vector(n+1, i, i!)==R_upto(n, 0)
test2(n)=vector(n+1, i, a(i))==R_upto(n, 1)

Is there a way to prove it? Is there a way to find exponential generating functions for $b(n,m)=R(n-1,0,m)$ with $b(0,m)=1$ and where $m\in\mathbb N$? Of course, here you may answer only the first question (and I am ready to choose such an answer as the correct one).