It is easy to show that $S_m=\sum_{n=1}^\infty \frac{n}{2^n + m}$ converges for any natural$\ m$, but what is its value?

In fact the series would converge even if$\ m$ were not natural, I just wanted to state that it is natural in my case. I have found the partial sum formula of$\ S_0$,$\displaystyle \sum_{n=1}^k \frac{n}{2^n} =\frac{2^{k+1}-k-2}{2^k}$, thus easily obtaining$\ S_0=2$. Then, since $\displaystyle \frac{1}{2^n+m}=\frac{1}{2^n}-\frac{m}{2^n\left(2^n+m\right)}$, I know $\displaystyle S_m= 2-\sum_{n=1}^\infty \frac{mn}{2^n\left(2^n+m\right)}$, though I'm not sure it is a convenient path to study the last series.

I think your observation helps if you iterate it more.

$$ \begin{align} S_m&=2-m\sum_{n=0}^\infty\frac{n}{2^n(2^n+m)}\\ &=2-m\sum_{n=0}^\infty \frac{n}{2^n}\left(\frac{1}{2^n}-\frac{m}{2^n(2^n+m)}\right)\\ &=2-m\sum_{n=0}^\infty \frac{n}{4^n}+m^2\sum_{n=0}^\infty\frac{n}{4^n(2^n+m)}\\ &=2-\frac49m+m^2\sum_{n=0}^\infty\frac{n}{4^n(2^n+m)}\\ \end{align}$$

Repeat this way, and brushing aside some convergence questions, we can write a power series in $m$:$$S_m=\sum_{n=0}^{\infty}(-1)^n\frac{2^{n+1}}{(2^{n+1}-1)^2}m^n$$ I'm not sure where you go from here, but you have an alternating power series now. Because it's alternating, it may be a quicker thing to use to get decimal approximations.

EDIT

Nope! This power series doesn't converge for $m\geq2$. However, with $m=2$, it's partial sums oscillate between two values, the average of which appears to be $S_2$, so that's interesting.

As a power series, this converges for $m\in(-2,2)$. You could at least use this to study $S_m$ for (the mostly non-integer) $m$ in $(-2,2)$.

I examined a graph of the power series on $(-2,2)$, and it looked similar to functions of the form $\frac{2\cdot2^r}{(m+2)^r}$. Experimenting, using an $r$ in the neighborhood of $0.455$ gives $\frac{2\cdot2^r}{(m+2)^r}$ that matches the power series fairly well, and appears to give a decent approximation for $S_m$. So perhaps you can more rigorously find an $r\approx0.455$ such that $S_m$ is asymptotically proportional to $\frac{1}{(m+2)^r}$.