Sure, the bottom of your kettle can get hotter than 212. Liquid won't, though. It will boil harder, but not hotter.
If it were that simple, then you'd never have to worry about scorching LME.
A few points to further complicate the conversation without offering anything more than mere conjecture.
Yes, liquid water will remain at its boiling point and no higher during the boil. Anything dissolved in said water would be at the same temperature as well. The thing is as soon as it becomes steam all bets are off and there is no maximum temperature. In addition not everything in the wort is dissolved. Larger particles are suspended and therefore could come in contact with said superheated steam directly.
Now, since the vast majority of the liquid (which posseses a rather large thermal mass) is set in stone in the ballpark of 212F, this superheated steam would rapidly come into equilibrium with the rest of the liquid as it rises through it.
Basically here's my opinion. There is a non-negligible chance that proteins suspended in wort could experience temperatures right at the hottest parts of the kettle to experience some Maillard reactions. But I have a tough time believing it would be a significant amount unless you're talking about a large, commercial-scale, direct-fire burner. Please note that this is a complete WAG, and I am not a physicist - I merely play one on the message boards.