I may have a different approach, provided that you know some cute things like Binomial Expansion and fractional summations.
Let's start with this (this will work for every $n$ and $m$):
$$z = \sin(x) ~~~~~ \text{d}z = \cos(x) = \sqrt{1 - \sin^2(x)} = \sqrt{1 - z^2}\ \text{d}x$$
Thus
$$\text{d}x = \frac{\text{d}z}{\sqrt{1 - z^2}} ~~~~~~~ \cos^n(x) = [\cos(x)]^n = \left(\sqrt{1 - \sin^2(x)}\right)^n = (1 - z^2)^{n/2}$$
Substituting into the integral and we get
$$\int z^m(1 - z)^{n/2}\frac{\text{d}z}{\sqrt{1 - z^2}} = \int z^m(1 - z^2)^{n/2 - 1/2}\ \text{d}z$$
Let's not call for simplicity $n/2 - 1/2 = \alpha$. Now we use the Binomial Expansion:
$$(1 - z^2)^{\alpha} = \sum_{k = 0}^{\alpha} \binom \alpha k (-z^2)^k\ \text{d}z$$
Thus substituting we get:
$$\sum_{k = 0}^{\alpha} \binom \alpha k (-1)^k\int z^{m+2k}\ \text{d}z$$
A trivial integration will lead us to
$$\sum_{k = 0}^{\alpha} \binom \alpha k (-1)^k \frac{z^{m+2k+1}}{m+2k+1}$$
Now substituting back for $z = \sin(x)$, and $\alpha = \frac{n-1}{2}$ we have
$$\boxed{\sum_{k = 0}^{\frac{n-1}{2}} \binom{\frac{n-1}{2}}{ k} (-1)^k \frac{\sin^{m+2k+1}(x)}{m+2k+1}}$$
Simple proof
Supposing we have $n = m = 1$, namely we are integrating $\int\sin(x)\cos(x)\ \text{d}x$ then we have
$$\frac{n-1}{2} = 0 ~~~~~~~ m + 2k + 1 = 1 + 0 + 1 = 2$$
thence
$$\int\sin(x)\cos(x)\ \text{d}x = \sum_{k = 0}^{0} \binom{0}{0} (-1)^0 \frac{\sin^{2}(x)}{2} = \frac{\sin^2(x)}{2}$$