Is $\mathbb{Z}[\sqrt{29}] $ a PID

Any UFD is normal, i.e. it equals its integral closure in its field of fraction. Here the integral closure of $ \mathbb{Z}[\sqrt{29}] $ in $ F $ contains the integral closure of $ \mathbb{Z}$ in $ F $, which is strictly larger than $ \mathbb{Z}[\sqrt{29}] $. Hence this ring is not normal, so it is not a UFD and therefore not a PID.