Does an ideal, frictionless sprinkler have a finite RPM? 

According to this fluid mechanics textbook, a sprinkler with torque generated by jets (the generator in the picture being disconnected) will have an finite equilibrium RPM.
I'm not sure if this is accurate. The argument is that once the velocity of the nozzle equals the velocity of the exiting water jet, the water is just falling out at no velocity relative to earth, and therefore not exerting any torque.
This seems hard to rationalize from the perspective of the nozzle, however; I think the nozzle is always experiencing "recoil force" no matter the RPM. Rockets routinely exert thrust when there exhaust velocity is lower than their flight velocity.
However, my professor agrees with the book.
Won't this sprinkler's rpm go up indefinitely? What am I missing here?
 A: If the sprinkler really were a rocket, in the sense that all the fuel was stored at the nozzle, then you would be absolutely right. In the frame of reference moving tangential to the rocket, it is pushing out fuel, which exerts a force on it. The exit velocity doesn't enforce a speed limit.
However whenever the sprinkler expels water, it also has to draw more water from pipe in the middle. That water starts out with zero angular momentum, and has to be accelerated to whatever angular momentum it has right before being shot out of the nozzle. Every action has an equal and opposite reaction, and so the sprinkler slows as the water moves outward, just like a spinning ballet dancer when they extend their arms.
To see how this cancellation works out to give the answer from your textbook, consider a drop of water of mass $m$ that starts out in the central pipe. It's initial angular momentum is $0$, and its angular momentum right before exiting the nozzle is $m\times r\times v_{tangential}$. By conservation of angular momentum, the sprinkler head loses this much angular momentum in the process. When the water droplet is shot out of the nozzle, its angular momentum is reduced by $m\times r\times v_{nozzle}$, and the sprinkler head gains this much angular momentum. Thus the total change in angular momentum of the sprinkler is $m\times r\times (v_{nozzle} - v_{tangential})$, which equals zero when $v_{nozzle} = v_{tangential}$. So it all works out.
