Finding a base of an exponent in mod field
Solving discrete logs modulo $2^n$ is easy.

You rely on this:

If we have (for odd $e$) $$(n 2^{k+1} + 2^k + 1)^e = b \pmod{2^\lambda}$$

then either:

$$(n 2^{k+1} + 2^k + 1)^e = b \pmod{2^{\lambda+1}}$$

or

$$(n 2^{k+1} + 2^k + 1)^{e + 2^{\lambda-k}} = b \pmod{2^{\lambda+1}}$$

That can be shown because $(n 2^{k+1} + 2^k + 1)^{2^{\lambda-k}} \equiv 1 + 2^\lambda \pmod {2^{\lambda+1}}$ (binomial theorem), and so $(n 2^{k+1} + 2^k + 1)^{e + 2^{\lambda-k}} \equiv (n 2^{k+1} + 2^k + 1)^e \cdot (1 + 2^\lambda) \equiv $ $(n 2^{k+1} + 2^k + 1)^e + 2^\lambda \pmod{ 2^{\lambda+1} }$

So, you can find your exponent by starting at the bottom, $9^1 = 9 \bmod 16$, and then extending it for increasing values of $\lambda$, until you hit $\lambda = 2048$