Problem 1348

1. Given the conditions:
- $m+1$ is a prime number of the form $4k-1$.
- There exists a prime number $p$ and a nonnegative integer $a$ such that:
$$\frac{m^{2^n-1}-1}{m-1} = m^n + p^a.$$

2. We start by rewriting the given equation:
$$\frac{m^{2^n-1}-1}{m-1} - m^n = p^a.$$
Let $D = \frac{m^{2^n-1}-1}{m-1} - m^n$.

3. We claim that $m^{2^k} + 1 \mid D$. To prove this, we need to show $m^{2^k} + 1 \mid mD$ since $\gcd(m, m^{2^k} + 1) = 1$.

4. Consider:
$$mD = \frac{m^{2^n} - 1}{m-1} - m^{n+1}.$$
Using the factorization of $m^{2^n} - 1$:
$$mD = (m+1)(m^2+1) \cdots (m^{2^{n-1}}+1) - (m^{2^k}+1)(m^{2^k(s-1)} - m^{2^k(s-2)} + \cdots).$$
This can be rewritten as:
$$mD = (m^{2^k}+1)\left((m+1)(m^2+1) \cdots (m^{2^{r-1}}+1)(m^{2^{r+1}}+1) \cdots (m^{2^{n-1}}+1) - m^{2^k(s-1)} - m^{2^k(s-2)} + \cdots\right).$$
Hence, $m^{2^k} + 1 \mid mD$.

5. Let $q = m+1$ be the prime. Then $q = m+1 \mid m^{2^k} + 1 \mid p^a$, which implies $p = q$.

6. Plugging $p = m+1$ into the initial equation, we get:
$$m^{2^n-2} + m^{2^n-3} + \cdots + m + 1 = m^n + (m+1)^a.$$

7. If $n \geq 3$, we consider two cases:
- If $m \equiv 2 \pmod{8}$:
$$m^{2^n-2} + m^{2^n-3} + \cdots + m + 1 \equiv 7 \pmod{8},$$
$$m^n + (m+1)^a \equiv 3^a \equiv 1, 3 \pmod{8}.$$
Hence, no solutions.
- If $m \equiv 6 \pmod{8}$:
$$m^{2^n-2} + m^{2^n-3} + \cdots + m + 1 \equiv 3 \pmod{8},$$
$$m^n + (m+1)^a \equiv 7^a \equiv 1, 7 \pmod{8}.$$
Hence, no solutions.

8. For $n = 2$, all $m+1$ such that $m+1$ is a prime of the form $4k-1$ works (here $a = 1$).

9. Therefore, the solutions are:
$$(m, n) = (p-1, 2),$$
where $p$ is a prime of the form $4k-1$.

The final answer is $\boxed{ (m, n) = (p-1, 2) }$ where $p$ is a prime of the form $4k-1$.