Power out = power in minus losses. Power out is 300 watts therefore power in will be about 330 watts. At 14 volts and 330 watts, current in will be 23.6 amps but this is average current. The peak current will depend on the duty cycle and looking at it simplistically, for a 50% duty cycle the peak current might be about 2 x 23.6 amps = 47 amps worst case (borderline discontinuous).
Ideally you need to pick an inductor that doesn't saturate and one with a gap is going to work better. A solid ferrite core will tend to start saturating at a flux density of about 0.4 teslas. It might have a relative permeability of (say) 1000. Using B = \$\mu H\$ allows this to be estimated: -
H = \$\dfrac{0.4}{4\pi \times 10^{-7} \times 1000}\$ = 318 A.t/m
You need to know the mean length of the magnetic field and that is found in the data sheet for the ferrite you are using. Let's say it is 100mm, this means your maximum ampere-turns is 31.8 and of course this would be unsuitable for your application because you can immediately see that it is likely your amperage and one turn is going to heavily saturate the core.
So, if you put a gap in it might reduce the effective permeability to (say) 200 and this allows more 5x H field (5x more current or 5x more turns) but inductance is related to turns squared so there is a net gain because although the \$A_L\$ value for the gapped core has gone down by a factor of 5 you only need \$\sqrt5\$ more turns to recover the inductance.
The S1JB-13-F diode has a pretty slow reverse recovery time (trr is 1.8us typ, 3us max). The boost switcher appears to be variable frequency, but at some loads it's switching frequency may be uncomfortably close (or right on top of) this reverse recovery time.
The MIC2251 data sheet recommends a Schottky diode but points out a fast switching diode can be used (their recommendation is the LS4148, whose trr is less than 8ns).
See this question for more on diode reverse recovery time.
A boost converter works by grounding the SW node during the 'on' time, which charges up the inductor. When the converter switches the internal FET off, the current flowing through the inductor needs to go somewhere, and the diode is preferred. So the voltage rises until the SW node is a diode drop above the desired output voltage. When the inductor has dumped its energy into the output cap/load, the diode needs to turn completely off. While it's turning off, you will get current flowing in the reverse direction, and depending on how long that recovery time is. The longer it is, the more time there is for current to flow backward and cause heating. In the worst case, if the switching frequency lines up badly, you can have the FET turning on again before the diode has recovered. I suspect that is what was happening in your case, since the converter was also exhibiting some temp rise and eventual failure.
You can see a couple examples of the operating frequency and switching waveforms in the graphs on Page 5 of the MIC2251 data sheet. And there is a graph of operating frequency vs load current on Page 4.
Best Answer
I wouldn't want to go above a peak flux density of 350 mT at 25 degC falling linearly to 300 mT if the core rises to 100 degC.
Now that depends on a few more things (as well as hysteresis loss implied by core saturation being kept reasonably low). So, work out what H field you will apply to avoid excessive flux density, add gapping if needed (it easily could be needed) and then work out what copper losses you might get with the extra turns (due to gapping) and take it from there.
If it's just core energy density you are considering, this is a guideline from the data sheet: -
But, to transfer this to your design you need to know the volume of your actual ferrite core and this isn't available in the material specification you posted a link to. Note that they are using 200 mT as the peak flux density.