You can rewrite your first sum as $\sum_n \frac{{n-1}\choose{m-1}}{n^s}$, because ${n-1}\choose{m-1}$ is the number of ways of writing $n$ as the sum of $m$ positive integers.
Since ${n-1}\choose{m-1}$ is a polynomial of degree $m-1$ in $n$, this series converges only when $\sum_n {n^{m-1-s}}$ converges, which is precisely when $s>m$.
Letting $q(n,m)$ be the number of ways of writing $n$ as the sum of $m$ positive squares, the second sum is $\sum_n \frac{q(n,m)}{n^{s/2}}$. So you'll need some estimate/bounds for $q(n,m)$ to figure out the values of $s$ for which this converges.
It's pretty easy to see that $q(n,m), for example, which shows convergence if $\frac{s}2>\frac{m}2 + 1$, but it seems likely that you'd have convergence for smaller $s$.
By coment below, since $\frac{1}{ m}(k_1+...+k_m)^2\leq k_1^2+...+k_m^2\leq (k_1+...+k_m)^2$, we see that if one of these series converges, then the other must, so the second series likewise converges exactly when $s>m$.