In an ideal transformer circuit, how is power transferred?
I initially thought that inductors won't store any power and the power
  must come from the mutual inductance. Am I correct in thinking that?


Yes, the equation is this for one inductor, \$\Phi \$ is the magnetic flux E is the EMF or voltage and N is the number of turns.

$$E= N\frac{d\Phi}{dt}$$

Since an ideal transformer has 100% mutual inductance connection from one coil to the next, it means that all of the magnetic flux is directly linked so \$\Phi \$ would be the same on both sides (in a real transformer some of the magnetic flux is lost and is not exactly the same on both sides)

$$E_{primary}= N_{primary}\frac{d\Phi}{dt} $$
$$E_{secondary}= N_{secondary}\frac{d\Phi}{dt} $$

Then you get this relationship
$$\frac{E_{primary}}{E_{secondary}}= \frac{N_{primary}}{N_{secondary}} $$

The power is transferred through the magnetic flux (not the magnetic field, because energy transfer can only happen if the magnetic field is changing hence the rate \$\frac{d\Phi}{dt}\$)

Inductors do store power and do so for a given amount of time, they do this by creating a magnetic field around the inductor as a current is converted to a magnetic field. If the current is removed, they generate voltage or EMF.

Transformers have a 'load' on their coil so they don't store energy as well as an inductor because the energy is transferred to the secondary coil.