The zero inflated Poisson regression model is defined for a sample $(y_1,\ldots,y_n)$ by $$ Y_i = \begin{cases} 0 & \text{with probability} \ p_i+(1-p_i)e^{-\lambda_i}\\ k & \text{with probability} \ (1-p_i)e^{-\lambda_i} \lambda_{i}^{k}/k! \end{cases}$$ and it further assumes that the parameters $\mathbf{\lambda} = (\lambda_1, \dots, \lambda_n)$ and $\textbf{p} = (p_1, \dots, p_n)$ satisfy
$$\eqalign{ \log(\mathbf{\lambda}) &= \textbf{B} \beta \\ \text{logit}(\textbf{p}) &= \log(\textbf{p}/(1-\textbf{p})) = \textbf{G} \mathbf{\gamma}. }$$
The corresponding log likelihood of the zero-inflated Poisson regression model is $$\eqalign{ L(\gamma,\mathbf{\beta}; \mathbf{y}) &= \sum_{y_i=0} \log(e^{G_i \gamma}+\exp(-e^{\textbf{B}_i \mathbf{\beta}})) +\sum_{y_i >0} (y_i \textbf{B}_i \mathbf{\beta}-e^{\textbf{B}_i \mathbf{\beta}})\\ &\quad -\sum_{i=1}^{n} \log(1+e^{G_{i} \gamma})-\sum_{y_i >0} \log(y_{i}!)}$$
Here, $\mathrm{B}$ and $\mathrm{G}$ are the design matrices. These matrices could be the same, depending on the features one desires to use for the two generating processes. They have the same number of rows, however.
Assuming that we could observe $Z_i = 1$ when $Y_i$ is from the perfect, zero state and $Z_i = 0$ when $Y_i$ is from the Poisson state the log-likelihood would be
$$L(\gamma,\mathbf{\beta}; \mathbf{y}, \mathbf{z}) = \sum_{i=1}^{n} \log(f(z_i|\mathbf{\gamma}))+\sum_{i=1}^{n} \log(f(y_i|z_i, \mathbf{\beta}))$$
$$ = \sum_{i=1}^{n} z_{i} (\textbf{G}_i \gamma-\log(1+e^{G_{i} \gamma}))+ -\sum_{i=1}^{n} (1-z_{i})\log(1+e^{G_{i} \gamma})+ \sum_{i=1}^{n} (1-z_i)[y_{i} \textbf{B}_i \beta-e^{\textbf{B}_i \beta} - \log(y_{i}!)]$$ The first two terms are the loss in a logistic regression to separate $z_i=0$ from $z_i=1$. The second term is a regression to the points generated by the Poisson process.
But aren't latent variables unobservable? The purpose is to maximize the first log-likelihood. But we have to introduce latent variables and derive a new log-likelihood. Then using the EM algorithm, we can maximize the second log-likelihood. But this assumes that we know that either $Z_i = 0$ or $Z_i = 1$?