Suppose we have the null and alternative hypotheses regarding a proportion
$H_0: p = p_0$
$H_a: p \ne p_0$
If we assume that $p$ takes the value $p_1$, what sample size do we need for a z-test of the null hypothesis at the $\alpha$ level with power $1 - \beta$?
The z-test based on the normal approximation rejects the null when the test statistic is greater than $z_{1 - \alpha/2}$,
$\big|{\frac{\hat{p} - p_0}{\sqrt{p_0 (1 - p_0) / n}}\big|} > z_{1-\alpha/2}$
Or, equivalently,
$\big|{\frac{\hat{p} - p_1}{\sqrt{p_1 (1 - p_1) / n}}\frac{\sqrt{p_1 (1 - p_1)}}{\sqrt{p_0 (1 - p_0)}} - \frac{p_1 - p_0}{\sqrt{p_0 (1 - p_0) / n}}\big|} > z_{1-\alpha/2}$
With $z = \frac{\hat{p} - p_1}{\sqrt{p_1 (1 - p_1) / n}}$, we have
$\big| z \frac{\sqrt{p_1 (1 - p_1)}}{\sqrt{p_0 (1 - p_0)}} - \frac{p_1 - p_0}{\sqrt{p_0 (1 - p_0) / n}}\big| > z_{1-\alpha/2}$
or equivalently,
$\big| z - \frac{p_1 - p_0}{\sqrt{p_1 (1 - p_1)/n}}\big| > z_{1-\alpha/2} \frac{\sqrt{p_0 (1 - p_0)}}{\sqrt{p_1 (1 - p_1)}}$
When $p = p_1$, $z \sim N(0,1)$ asymptotically, and the asymptotic power of the test is therefore equal to
$1 - \beta = \Phi\left( \frac{\sqrt{n}|p_1 - p_0| - z_{1-\alpha/2}\sqrt{p_0 (1 - p_0)}}{\sqrt{p_1 (1 - p_1)}} \right) + \Phi\left( \frac{-\sqrt{n}|p_1 - p_0| - z_{1-\alpha/2}\sqrt{p_0 (1 - p_0)}}{\sqrt{p_1 (1 - p_1)}} \right)$
Typically the second term on the right of the equation is quite small, and so we can approximate
$1 - \beta = \Phi\left( \frac{\sqrt{n}|p_1 - p_0| - z_{1-\alpha/2}\sqrt{p_0 (1 - p_0)}}{\sqrt{p_1 (1 - p_1)}} \right)$
Taking the inverse of the normal on both sides,
$z_{1-\beta} = \frac{\sqrt{n}|p_1 - p_0| - z_{1-\alpha/2}\sqrt{p_0 (1 - p_0)}}{\sqrt{p_1 (1 - p_1)}} $
Solving for $n$,
$n = \left(\frac{z_{1-\beta}\sqrt{p_1 (1 - p_1)} + z_{1-\alpha/2}\sqrt{p_0 (1 - p_0)}}{p_1 - p_0}\right)^2 $
One approximation to the power function is to use the p-value function. The p-value for a score test of $H_0: p\le p_0$ vs $H_a: p > p_0$ is
$$H(p_0;\hat{p})=1-\Phi\bigg(\frac{\hat{p}-p_0}{\sqrt{p_0(1-p_0)/n}}\bigg).$$
At the minimum detectable effect $\hat{p}_{MDE}$ the p-value testing $H_0:p\le p_0$ equals the alpha level of the test, $H(p_0;\hat{p}_{MDE})=\alpha$. For a given $n$ this p-value function evaluated at any other value $p$ is approximately equal to the power of the test,
$$1-\beta(p)\approx H(p;\hat{p}_{MDE})=1-\Phi\bigg(\frac{\hat{p}_{MDE}-p}{\sqrt{p(1-p)/n}}\bigg),$$
where the minimum detectable effect is $\hat{p}_{MDE}=p_0+z_{1-\alpha}\sqrt{p_0(1-p_0)/n}$.
This can be numerically solved for the sample size $n$. This approach can be simplified even further by inverting a Wald test,
$$1-\beta(p)\approx H(p;\hat{p}_{MDE})=1-\Phi\bigg(\frac{\hat{p}_{MDE}-p}{\sqrt{\hat{p}_{MDE}(1-\hat{p}_{MDE})/n}}\bigg),$$
where the minimum detectable effect is $\hat{p}_{MDE}=p_0+z_{1-\alpha}\sqrt{\hat{p}_{MDE}(1-\hat{p}_{MDE})/n}$. This is particularly useful in more complicated model. Of course for a binomial proportion the power of a test can be solved exactly by identifying the rejection region of the test and then evaluating the complementary CDF of the binomial distribution under the alternative.
Here is a paper discussing inference on power as well as approximating power using a p-value function.
