I think of Zaitsev elimination reactions a SN1-like. They occur from elimination from electron donation of the most electron rich carbon, that is the most substituted. I think of Hoffman elimination products as the result of deprotonation of the most acidic hydrogen, this will be the least substituted.
The base, solvent, and leaving groups can all affect these elements. A very good leaving group, like iodide will stretch the C-I bond and draw electron density toward it. This will favor Zaitsev, as will a polar solvent. A poor leaving group or one with a higher pKa, will not draw electrons in the same manner. A fluoride or quaternary ammonium salt are more basic, poorer leaving groups, pull electrons from neighboring groups to a lesser degree, and are dependent upon deprotonation initiating the reaction. The least substituted carbon is the most acidic hydrogens. This will give you a Hoffman product.