This answer is half correct. Even if the HI protonated the phenol, which would convert it into a good leaving group, the iodide anion could not attack in a SN2 reaction since it could not undergo a backside attack on the ring carbon. The protonated phenol would not be a good enough leaving group to actually leave the benzene ring and form a carbocation on the ring carbon because the geometry of the ring would make this particular carbocation very unstable.

However the alcohol is protonated and readily undergoes a SN2 reaction to form the benzyl iodide.