For the second equality recall the conbinatorial species for factorizations into cycles which is

$$\mathfrak{P}(\mathfrak{C}(\mathcal{Z}))$$

which yields

$$\left[k\atop m\right] = k! [z^k] \frac{1}{m!} \left(\log\frac{1}{1-z}\right)^m.$$

This yields for the RHS of identity two

$$\frac{n!}{m!} \sum_{k=m}^n [z^k] \left(\log\frac{1}{1-z}\right)^m = \frac{n!}{m!} [z^n] \frac{1}{1-z} \left(\log\frac{1}{1-z}\right)^m = Q_{n,m}.$$

We also have

$$\sum_{n\ge m+1} \left[n\atop m+1\right] \frac{z^n}{n!} = \frac{1}{(m+1)!} \left(\log\frac{1}{1-z}\right)^{m+1}$$

which implies that (differentiate)

$$\sum_{n\ge m+1} \left[n\atop m+1\right] \frac{z^{n-1}}{(n-1)!} = \frac{1}{m!} \left(\log\frac{1}{1-z}\right)^m \times (1-z) \times \frac{1}{(1-z)^2}$$

or alternatively

$$\sum_{n\ge m} \left[n+1\atop m+1\right] \frac{z^{n}}{n!} = \frac{1}{m!} \frac{1}{1-z} \left(\log\frac{1}{1-z}\right)^m$$

which means that

$$Q_{n,m} = \left[n+1\atop m+1\right].$$

For the second equality recall the conbinatorial class for factorizations into cycles which is

$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \textsc{SET}(\textsc{CYC}(\mathcal{Z}))$$

which yields

$$\left[k\atop m\right] = k! [z^k] \frac{1}{m!} \left(\log\frac{1}{1-z}\right)^m.$$

This yields for the RHS of identity two

$$\frac{n!}{m!} \sum_{k=m}^n [z^k] \left(\log\frac{1}{1-z}\right)^m = \frac{n!}{m!} [z^n] \frac{1}{1-z} \left(\log\frac{1}{1-z}\right)^m = Q_{n,m}.$$

We also have

$$\sum_{n\ge m+1} \left[n\atop m+1\right] \frac{z^n}{n!} = \frac{1}{(m+1)!} \left(\log\frac{1}{1-z}\right)^{m+1}$$

which implies that (differentiate)

$$\sum_{n\ge m+1} \left[n\atop m+1\right] \frac{z^{n-1}}{(n-1)!} = \frac{1}{m!} \left(\log\frac{1}{1-z}\right)^m \times (1-z) \times \frac{1}{(1-z)^2}$$

or alternatively

$$\sum_{n\ge m} \left[n+1\atop m+1\right] \frac{z^{n}}{n!} = \frac{1}{m!} \frac{1}{1-z} \left(\log\frac{1}{1-z}\right)^m$$

which means that

$$Q_{n,m} = \left[n+1\atop m+1\right].$$