Thumb didn't work. Oh well.

Ah, found the thing. [link]

I'm thinking it's finite.

Let M_n = 999... [n 9's] ...999,
Let d(n) = floor(log(n)) + 1 ; representing the number of digits in n.
Let f(n) = the sum of the digits each raised to the power of d(n).
We will say it "has the property" if f(n) = n for some n.

If d(f(M_n)) < n for some n, there does not exist a number satisfying the property for that n. (If M doesn't satisfy it, nothing will.)

f(M_n) = n*9^n
d(f(M_n)) = floor(log(n*9^n)) + 1
= floor(log(n) + log(9^n)) + 1
= floor(log(n) + n*log 9) + 1
<= log(n) + n*log 9 + 1 because nobody likes the floor function when we're trying to algebra.

log 9 = 0.95424250943932487459005580651023 according to windows XP's calculuator which should really allow copy and paste.

Suppose n = 100. Then log(100) + 100*log 9 + 1 = 2 + 95.4242... + 1 = 98.4242 which is less than 100. So there are no hundred-digit numbers with this property.

So by induction, since
log(n) + n*log 9 + 1 < n
is true for n = 100, I want to show that
log(n+1) + (n+1)*log 9 + 1 < n + 1.

When n > 100, we have:
log(1 + 1/n) < log (1 + 1/9)
log((n+1)/n) < log (10/9)
log(n+1) - log n < log 10 - log 9
log(n+1) + log 9 < log n + 1
log(n+1) + n*log 9 + log 9 < log n + n*log 9 + 1 < n (given in the base step)
log(n+1) + n*log 9 + log 9 < n
log(n+1) + (n+1)*log 9 < n
log(n+1) + (n+1)*log 9 + 1 < n+1

I'm not sure induction was even necessary here. Still, this shows that when n > 100, there does not exist a number with the property. So if this is all sound, the set of numbers must be finite.

Now you've got me curious! I'm going to try and get a program to list all the numbers before the night is out.