, 2003

Solution:

First of all, $n$ must be odd in both cases, because $2^{m}$ is even for $m \geq 1$.

i. $n = m = 3$. Indeed, among two consecutive even numbers one is not divisible by four, while from $2^{m} = n^{2} - 1 = (n-1)(n+1)$ we obtain that $n-1$ and $n+1$ are both powers of 2; hence the only remaining possibility is $n-1 = 2$, from which $n = 3$ and therefore $m = 3$.

Alternatively, note that, setting $n = 2a + 1$, we have $n^{2} - 1 = 4a^{2} + 4a = 4a(a+1)$, which is possible only for $a = 1$, otherwise $2^{m}$ would have an odd factor greater than 1.

ii. $n = m = 1$. Setting $n = 2a + 1$ we obtain $2^{m} = 4a^{2} + 4a + 2$, which gives remainder 2 when divided by four. Hence $m$ must be one, otherwise $2^{m}$ would be divisible by 4, and $a$ must therefore be 0. On the other hand, it is immediately seen that $n = m = 1$ is a solution $(2^{1} - 1 = 1)$.