Actually, Bernstein's inequality does not really require boundedness of the i.i.d. random summands; a finite exponential moment of the absolute value of a random summand will suffice. However, here we can just use Markov's inequality. 

Let $X,X_1,\dots,X_n$ be independent identically distributed random variables (i.i.d. r.v.'s) such that $P(X=z)=\mu(z)\in(0,1)$ for all $z$ in a finite set $Z$, with $\sum_z\mu(z)=1$. Let $Y:=-\ln\mu(X)$, $Y_i:=-\ln\mu(X_i)$, and $S_n:=\frac1n\,\sum_1^n Y_i$. Then 
$$ES_n=EY=-\sum_z\mu(z)\ln\mu(z)=:H(\mu)>0  
$$ 
and 
$$Ee^{hY}=\sum_z\mu(z)^{1-h},\quad Ee^{hS_n}=(Ee^{hY/n})^n \tag{1}
$$
for all real $h$. Note also that 
$$-\max_z\ln\mu(z)=\min Y<EY=H(\mu)<\max Y=-\min_z\ln\mu(z).$$   

Take now any real $t$ such that $H(\mu)=EY\le t<\max Y$. For all real $h\ge0$, by Markov's inequality, 
$$P(S_n\ge t)\le\exp\{-ht+\ln Ee^{hS_n}\}=\exp\{-ht+n\ln Ee^{hY/n}\}. \tag{2}
$$
The derivative of the exponent $-ht+n\ln Ee^{hY/n}$ in $h$ is $-t+\frac{EYe^{hY/n}}{Ee^{hY/n}}$, which strictly and continuously increases from $-t+EY\le 0$ to $-t+\max Y\ge0$ as $h$ increases from $0$ to $\infty$, and so, the upper bound $\exp\{-ht+n\ln Ee^{hY/n}\}$ on the right-tail probability $P(S_n\ge t)$ is minimized when $h=h_{t,+}$ is the only nonnegative root of the equation 
$$\frac{EYe^{hY/n}}{Ee^{hY/n}}=t. \tag{3} 
$$

Similarly, for any real $t$ such that $H(\mu)=EY\ge t>\min Y$, the best upper exponential bound on the left-tail probability $P(S_n\le t)$ is $\exp\{-ht+n\ln Ee^{hY/n}\}$, where now $h=h_{t,-}$ is the only non-positive root of the equation $(2)$. 

In view of $(1)$, equation $(3)$ can be rewritten as 
$$\sum_z(t+\ln\mu(z))\mu(z)^{1-s}=0,  \tag{4}
$$
where $s:=h/n$, 
and this equation can be easily solved for $s$ numerically if the set $Z$ is not too large.  

Note that then the upper bound in $(2)$ can be rewritten as $e^{-na}$, where 
$a:=st-\ln Ee^{sY}>0$ if $t\ne EY$, so that this bound exponentially decreases in $n$ if $t$ is fixed. 

All this is of course well known, even in the general case of i.i.d. random summands with finite exponential moment of the absolute value. Basically, in this particular situation I have just added the first equality in $(1)$ and rewrote $(3)$ as $(4)$.