2^x mod 30 = 1, 2, 4, 8, or 16

5^y mod 30 = 1, 5, or 25

31^z mod 30 = 1

Then 2^x + 5^y - 31^z mod 30 = 1, 2, 4, 5, 6, 8, 10, 12, 16, 20, 25, 26, or 28

n! mod 30 = 0, 1, 2, 6, or 24

The overlap of congruences of the two sides of the equation is 1, 2, or 6. Then n! can only be 1, 2, or 6. Larry's brute force search finds all three values of n! are possible.

What has not been proven yet is that there are not any solutions with "large" values for x, y, and z.