The set of such $n$ is infinite. In fact, note that modulo $67$, the possible residues of $33$-rd powers are just $0$, $1$ and $-1$ by Fermat. Thus, to express a number in the congruence class $12 \pmod{67}$ as a sum of $33$-rd powers we need at least $12$ summands. On the other hand, of course infinitely many numbers in this congruence class can indeed be expressed as sums of twelve $33$-rd powers, e.g. $(67k+1)^{33}+11$ for any $k$.

The set of such $n$ is finite, in fact empty. Indeed, let us show that any $n$ has $b(n) \le 12^{65}$ (which could be much strengthened, but completely suffices for us). The main point is taking finite differences to reduce to a linear polynomial. E.g. with $(x+1)^{33}-x^{33}=33x^{32}+\dots$ we get a degree-$32$-polynomial by taking a suitable sum of two $33$-rd powers. Iterating, we find that $\Delta^{32}(x^{33})$ is a linear polynomial $33!x+c$ and can be described as a sum of $2^{32}$ terms of the form $(x+k)^{33}$. Thus, we can obtain any number in a fixed congruence class modulo $33!$ with at most $2^{32}$ terms, and then of course we can just add some ones to reach each number with at most $2^{32}+33! <12^{64}+12^{64}<12^{65}$ summands.