Let $Y$ be random variable with $Poiss(\theta)$ distribution. The parameter $\theta$ is a realization of a random variable $\Theta$ with a priori distribution $Exp(\lambda)$.
The task is to find HPD region (shortest credible interval) for $\Theta$ and $1-\alpha = 0.95$ if we observed $Y=0$.
$\Theta|Y=0 \sim Exp(\alpha + 1)$.
I want to find $q$ such that: $P(\Theta>q|Y=0) = 1 - \lambda$
$e^{-(\lambda +1)q} = 1 - \lambda$, so $q = - \frac{ln(1-\lambda)}{\lambda+1}$.
Am I right? What to do next?
I cannot follow your calculations. Mine are:
You start with $f_\Theta(\theta)=e^{-\theta}$ and $\mathbb P(Y=0 \mid \Theta=\theta)=e^{-\theta}$ so $f_{\Theta \mid Y=0}(\theta) = 2e^{-2\theta}$ when $\theta \gt 0$
This posterior density is a decreasing function of $\theta$ and so the $95\%$ credible interval starts at $0$ and ends at $- \log_e(0.05) / 2 \approx 1.497866$