Solution:
Set $m = 2^{k} p$, where $p > 1$ is an odd integer. Then
$$2003^{m} - 1 = 2003^{2^{k} p} - 1 = \left(2003^{2^{k}} - 1\right) K$$
where $K$ is a sum of $p$ even integers and $1$; in particular, $K$ is odd. Hence $m$ must be of the form $m = 2^{k}$. In this case we have that
$$\begin{aligned} 2003^{2^{k}} - 1 & = \left(2003^{2^{k-1}} - 1\right)\left(2003^{2^{k-1}} + 1\right) \\ & = \left(2003^{2^{k-1}} + 1\right)\left(2003^{2^{k-2}} + 1\right) \ldots (2003 + 1)(2003 - 1) \end{aligned}$$
Since $2003^{2^{i}} + 1 \equiv 2 \pmod{4}$, then $2^{k+2}$ divides $2003^{2^{k}} - 1$ but $2^{k+3}$ does not (use that $2003 + 1 \equiv 4 \pmod{8}$ and $2003 - 1 \equiv 2 \pmod{4}$). Therefore $k + 2 = 2000$, i.e. $k = 1998$. Thus the desired number is equal to $2^{1998}$.