The Euler product for the Riemann zeta function $\zeta(z)$ implies that $$ \log\zeta_R(z)=\sum_{m>0}\frac{P(mz)}{m} \tag{R}, $$ whereas the Ihara zeta function for a graph $G$--all can be expressed in the same basic form:
$$\log \zeta_I(z)= \sum_{m> 0} \frac{N_mz^m}{m} .\tag{I}$$
What confuses me is the different appearance of the variable $z$. Since $\displaystyle P(mz)=\sum_{p\,\in\mathrm{\,primes}} (p^{-z})^m$, we have $z$ in the exponent in $(R)$ , but polynomial in $(I)$.
Why is that?
EDIT: Further I found the following on The Riemann hypothesis for ($q+1$-regular) graphs: $$\displaystyle \zeta(s) = \prod_{p \ \rm prime} \frac { 1 } { 1 - p^{-s}} \ \ \ \ \ (2), $$ compared to $$ \displaystyle z(s) = \prod_{p \ \rm prime\ cycle} \frac 1 { 1 - (\color{red}{q^{|p|}})^{-s }} \ \ \ \ \ (5) $$ which still looks confusing to me since we build a product over primes in Riemann's case, where the prime $p$ varies, while a product over a fixed variable $\color{red}q$ with the length of the prime cycle as an exponent in Ihara's case, but since $(5)$ would just use powers of $q$, like as I would restrict my set of primes to only one single prime in $(2)$ it might be OK...