For $a\ge 1$, the quantity $(m+k+1)^a-m^a$ is increasing w.r.to $m\ge0$ (because $x\mapsto x^a$ is convex). So for $0<b<1$, the term $b^{(m+k+1)^a-m^a} $ is decreasing w.r.to $m\ge0$, and so is $\sum_{k\ge0}b^{(m+k+1)^a-m^a}=b^{-m^a} \sum_{k> m}b^{k^a}$. That is, we have
the required bound with $$c=c(b):= \sum_{k>0} b^{k^a} , $$ which is optimal as it gives an equality for $m=0$ .

For $0<a<1$ the answer is already answered in negative by Christian Remling.

(Note that if $0<a<1$, the above inequality are to be inverted, by concavity of $x\mapsto x^a$, and $ b^{-m^a} \sum_{k> m}b^{k^a}$ is increasing; moreover $(m+k+1)^a-m^a\le am^{a-1}
(k+1)$ so $$b^{-m^a} \sum_{k> m}b^{k^a}=\sum_{k\ge0}b^{(m+k+1)^a-m^a}\ge \sum_{k\ge0}b^{am^{a-1}
(k+1)}=\frac{1}{1-b^{ {a}{m^{a-1}}} }-1,$$
that diverges like $m^{1-a}$ as $m\to+\infty$.)