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
\(V\left(b_1|\beta_1=\theta\right)\)
is the variance of the estimate \(b_1\) given the true slope
\(\beta_1=\theta\).
The variances are
$$
V\left(b_1 | \beta_1 = 0\right)=\frac{1}{\sigma^2_{x_1}}
$$
and
$$
V\left(b_1 | \beta_1 = \theta_1\right)=\frac{1}{\sigma^2_{x_1}}
\exp\left[-\left(\theta_1 \mu_{x_1} + \theta_1^2\sigma^2_{x_1}/2\right)\right]
$$