Sum of digits of x and powers of x

[mart_r]

While this thread is still on the front page, let me drop this number here:

977501592599998899778899999999987869899997877899798979989989676889999889999899775899989798999998789769998994897988599999999988989899879989699897999798898997599889876898998999994977897978999988999939998887

It's a 204-digit prime number whose digital sum (1651) is larger than the digital sum of its cube (1639). I found it less than a day after my last post in this thread.

[mart_r]

This could make for a nice puzzle, or at least serve as inspiration for further endeavors:
(assume all mentioned variables being positive integers)

For any base b, are there only finitely many numbers x not divisible by b such that the sum of digits of x is larger than the sum of digits of \(x^5\) (in base b)?
Or is there a threshold \(b_0\) above which all \(b > b_0\) can have infinitely many (or at least one) such solutions? (Is \(b_0\)=283 for fifth powers?)
Some solutions for \(b \leq 100\) (searched up to \(x=10^8\)):

 b  x
 8  4*
27  9*, 23
32  2*, 4*, 8*, 16*
39  177716
40  20*
53  8210
54  18*, 36*, 138*
55  31
60  42
64  16*, 32*, 48*, 245408*
72  36*
77  822816
79  16255431
90  299047
92  52881676
96  24*, 48*, 6600*, 17256*
98  7140*
* semi-trivial solutions, since b|x^5

For \(b > 100\), more and more non-trivial solutions appear (the next one is b=102, x=4767).
b=27 seems to yield the smallest non-trivial solution. Or might there be a smaller b for which such a solution can be found?

What about higher prime powers p? It is trivial to show that there are arbitrarily large bases b for which \(x \geq 3\) (mutually, all \(x^n\) for \(1 \leq n \leq p-1\)) is a solution whenever \(b=x^p+2-x\). Excluding all those trivial and semi-trivial solutions, for p=7 so far the smallest (in terms of b) non-trivial solution I found was b=492, x=121820.
Is it possible to find a non-trivial solution for larger p? Have I overlooked a way to trivially construct solutions?