As a flyback works by storing energy and then releasing it again, is
my assumption correct that this transformer could store more energy
due to the higher inductance and theoretically allow a higher output
Your assumption is incorrect.
- The primary inductor stores only so-much energy in a given time.
- That time is determined by the duty cycle and the operating frequency.
So, when you connect the primary to the DC supply with a MOSFET, the primary current ramps up at V/L amps per second. If L increases then, the current ramps up at a slower rate and, by the time the MOSFET has deactivated, the peak current will be smaller.
Given that energy stored is proportional to current squared you can see that there is a net loss with this idea despite the inductance being greater. Basically, doubling the inductance halves the energy stored per cycle. You would have to operate at a lower frequency to make things equal.
As I don't need such a high output power...
250 watts is a lot for a flyback converter.
I assume the smaller core would work and would cause less core losses?
No, not normally. The core you chose is this one (the JA4635) and, it has a maximum primary current of 10.5 amps so, with your peak current of 17 amps (from comments), it would heavily saturate.
Heavy saturation means that instead of the primary current rising linearly, it rapidly shoots off to a very much higher value and you end up in deeper saturation with the possibility of destroying the MOSFET that tries to control current.
Or, if the control circuit does its job well then you can regard excessive saturation as producing a much lower value of inductance, so you don't get the energy storage you want and you don't meet your specification.