a) Without loss of generality, we may suppose $1 \le m < n$.
If $m$ and $n$ are odd, then $m \ge 1$ and $n \ge 3$. Taking $d_1 = 1$ and $d_2 = n$ yields $d_1 + d_2 = n + 1$, which is an even number at least $4$, so it is composite, contradiction.
If $m$ and $n$ are even, taking $d_1 = 2$ and $d_2 = 2$ yields $d_1 + d_2 = 4$, contradiction.
In conclusion $m$ and $n$ have different parities, so $m + n$ is odd.

b) If $m = n$, then $d_1 = m$ and $d_2 = n$ leads to $d_1 + d_2 = 2m$, which must be a prime, hence $m = n = 1$.
If $m < n$, then $m \neq n$ and, from a), $m$ and $n$ have different parities.

* If $m$ is even and $n$ is odd, then $m \ge 2$ and $n \ge 3$. Then $d_1 = 1$ and $d_2 = n$ yields $d_1 + d_2 = n + 1 \ge 4$, contradiction.
* If $m$ is odd and $n$ is even, then $m \ge 1$ and $n \ge 2$. Take $a \in \mathbb{N}^*$ and $b$ odd such that $n = 2^a \cdot b$.

I. If $a \ge 3$, then $8$ divides $n$ and, taking $d_1 = 1$ and $d_2 = 8$ yields $d_1 + d_2 = 9$, contradiction.

II. If $a = 1$, then $n = 2b$, with odd $b$.

i) If $b \ge 3$, then take $d_1 = 1$ and $d_2 = b$ to get $d_1 + d_2 = b + 1 \ge 4$, contradiction.

ii) If $b = 1$, then $n = 2$ and, since $m < n$ and $m$ is odd, $m = 1$. It is easy to check that the pair $(m, n) = (1, 2)$ is a solution.

III. If $a = 2$, then $n = 4b$, with odd $b$.

i) If $b \ge 3$, then $d_1 = 1$ and $d_2 = b$ yields $d_1 + d_2 = b + 1 \ge 4$, contradiction.

ii) If $b = 1$, then $n = 4$ and, since $m < n$ and $m$ is odd, $m \in \{1, 3\}$. It is easy to check that only $(m, n) = (1, 4)$ satisfies the statement.

Finally, the solutions are $(1, 1)$, $(1, 2)$, $(1, 4)$.