Let $m$ be any positive integer. What is the exponent of 2 in the prime factorization of $1991^{m}-1$?
Let the exponent of 2 in the prime factorization of the number $1991^{m}-1$ be denoted by $f(m)$. If $m=2^{k} \cdot p$, where $p$ is an odd number, then $\left(1991^{m}-1\right)$ can be factored as the difference of two $p$-th powers, with the first factor being $\left(1991^{2^{k}}-1\right)$ and the second factor being odd, so $f(m)=f\left(2^{k}\right)$, where $k$ is the exponent of 2 in the prime factorization of $m$.

Now, $1991^{2^{k}}-1$ can be factored as follows:

$$
\left(1991^{2^{k-1}}+1\right)\left(1991^{2^{k-2}}+1\right) \ldots(1991+1)(1991-1)
$$

If $r>0$ and is even, then $1991^{r}+1$ is divisible by 2 but not by 4, since $1991^{r}+1=(1992-1)^{r}+1$ gives a remainder of 2 when divided by 4. Since $1992=2^{3} \cdot 249,1990=2 \cdot 995$, it follows from the above that

$$
f\left(2^{k}\right)=\left\{\begin{array}{l}
1, \quad \text { if } k=0 \\
k+3, \text { if } k>0
\end{array}\right.
$$

This means that if $k$ represents the exponent of 2 in the prime factorization of $m$, then for an odd exponent $m$ (i.e., $k=0$), the value of the desired $f(m)$ exponent is 1, and otherwise it is $k+3$.