The simple Poisson regression has the following form:
$$
Pr(Y_i = y_i | mu_i, t_i) = \exp(-\mu_i t_i) (\mu_i t_i)^{y_i}/ (y_i!)
$$
where
$$
\mu_i=\exp(\beta_0+\beta_1 x_{1i})
$$
We are interested in testing the null hypothesis $\beta_1=0$
versus the alternative hypothesis $\beta_1 = \theta_1$.
Assume $x_{1}$ is normally distributed with mean
$\mu_{x_1}$ and variance $\sigma^2_{x_1}$.
The sample size calculation formula derived by Signorini (1991) is
$$
N=\phi\frac{\left[z_{1-\alpha/2}\sqrt{V\left(b_1 | \beta_1=0\right)}
+z_{power}\sqrt{V\left(b_1 | \beta_1=\theta_1\right)}\right]^2}
{\mu_T \exp(\beta_0) \theta_1^2}
$$
where $\phi$ is the over-dispersion parameter ($=var(y_i)/mean(y_i)$),
$\alpha$$ is the type I error rate,
$b_1$ is the estimate of the slope $\beta_1$,
$\beta_0$ is the intercept,
$\mu_T$ is the mean exposure time,
$z_{a}$ is the $100*a$-th lower percentile of
the standard normal distribution, and $V\left(b_1|\beta_1=\eta\right)$
is the variance of the estimate $b_1$ given the true slope
$\beta_1=\eta$.
The variances are
$$
V\left(b_1 | \beta_1 = 0\right)=\frac{1}{\sigma^2_{x_1}}
$$
and
$$
V\left(b_1 | \beta_1 = \theta_1\right)=\frac{1}{\sigma^2_{x_1}}
\exp\left[-\left(\theta_1 \mu_{x_1} + \theta_1^2\sigma^2_{x_1}/2\right)\right]
$$
References
Signorini D.F. (1991).
Sample size for Poisson regression.
Biometrika. Vol.78. no.2, pp. 446-50