$\newcommand\ep\epsilon$Let $X_i:=c^{i-1}(x_i-Ex_i)$, so that the $X_i$'s are independent zero-mean random variables. The condition $c=1+1/m$ for $m\ge n$ implies that 
$$c=1+b/n,$$
where $0<b=O(1)$. Let $S:=\sum_1^n X_i$. We have to upper-bound $P(S\ge\ep)$ for real $\ep>0$. 

It follows from the proof of inequality (2.9) in [Hoeffding (1963)][1] (see formula (4.18) in Hoeffding's paper and the last equality in formula (12) in [Bennett (1962)][2]) that 
$$ P(S\ge\ep)\le Q(\ep):=\exp\Big\{\frac{B^2}{y^2}\,\psi\Big(\frac{\ep y}{B^2}\Big)\Big\},\tag{10}\label{10}$$
where $\psi(u):=u-(1+u)\ln(1+u)$, and $B^2$ and $y$ are any positive real numbers such that 
$$X_i\le y\text{ for all }i\text{ and }\sum_1^n EX_i^2\le B^2. \tag{20}\label{20} $$
(Inequality (2.9) in Hoeffding's paper was established using a condition that can be written, in our terms, as $\sum_1^n EX_i^2=B^2$, instead of the condition $\sum_1^n EX_i^2\le B^2$ in \eqref{20}. 
A much simpler way to derive \eqref{10} assuming \eqref{20} is to note that the function $r$ is increasing on $\Bbb R$, where $r(u):=(e^u-1-u)/u^2$ for real $u\ne0$ and $r(0):=1/2$. -- See the details at the end of this answer.)

Note that 
$$X_i\le c^{i-1}\le c^n=(1+b/n)^n<e^b$$
for all $i=1,\dots,n$ and 
$$\sum_1^n EX_i^2=\sum_1^n c^{2i-2}\frac1n\Big(1-\frac1n\Big)
\le\frac1n\frac{c^{2n-1}-1}{c^2-1}<\frac{e^{2b}-1}{2b}.$$
So, \eqref{10} holds with 
$$B^2=\frac{e^{2b}-1}{2b},\quad y=e^b.$$

Note that, in view of the condition $0<b=O(1)$, we have $B^2\asymp1$ and $y\asymp1$. So, the bound $Q(\ep)$ in \eqref{10} will go to $0$ iff $\ep\to\infty$, and then we will have 
$$Q(\ep)=e^{-C\ep\ln\ep},$$
where $C\asymp1$, so that the distribution of $S$ has a Poisson-like right tail. This shows that the bound $Q(\ep)$ on $P(S\ge\ep)$ is good, because even for $b=0$ the distribution of $S$ converges to a Poisson distribution (as $n\to\infty$). 

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*Proof of \eqref{10}:* For any real $h\ge0$, 
\begin{align}
P(S\ge\ep)&\le e^{-h\ep}\prod_1^n Ee^{hX_i} \\ 
&\le e^{-h\ep}\exp\sum_1^n (Ee^{hX_i}-1) \\ 
&= e^{-h\ep}\exp\sum_1^n (Ee^{hX_i}-1-hX_i) \\ 
&= e^{-h\ep}\exp\sum_1^n r(hX_i)h^2EX_i^2 \\ 
&\le e^{-h\ep}\exp\sum_1^n r(hy)h^2EX_i^2 \\ 
&\le e^{-h\ep}\exp(r(hy)h^2B^2) \\ 
&= \exp\{-h\ep+(e^{hy}-1-hy)B^2/y^2\} \\ 
&=Q(\ep)
\end{align}
if $h=\frac1y\,\ln(1+\frac{\ep y}{B^2})$ (the minimizer of $-h\ep+(e^{hy}-1-hy)B^2/y^2$ in $h$). $\quad\Box$

  [1]: http://www.jstor.org/stable/2282952?origin=JSTOR-pdf
  [2]: http://www.jstor.org/stable/2282438?origin=JSTOR-pdf