Even if the answer of @Chris is right, I'd like to share mine :)

To check the power p of a number n, you can convert this number to the base p and check if the first digit (on the left) is 1, and the remaining digits are 0.

x = 3**8
y = x - 1
z = x - 9
a = 30

def ternary (n):
    if n == 0:
        return '0'
    nums = []
    while n:
        n, r = divmod(n, 3)
        nums.append(str(r))
    return ''.join(reversed(nums))

def is_power_of_3(n):
    nstr = ternary(n)
    return nstr[0] == '1' and int(nstr[1:]) == 0

print("x = {} {}\ny = {} {}\nz = {} {}\na = {} {}".format(ternary(x), is_power_of_3(x), ternary(y), is_power_of_3(y), ternary(z), is_power_of_3(z), ternary(a), is_power_of_3(a)))

I gave 4 example, only the x is a power of 3

Output:

x = 100000000 True
y = 22222222 False
z = 22222200 False
a = 1010 False