For $k=1$ each problem was solved by only one student. That means that there are no other students, due to the condition that for any two students there is exactly one problem which they both solved.

Now let $k \ge 2$.
Let $A$ be one of the students. He solved exactly three problems, but each of these three problems was solved by $k-1$ more students. However, these $3(k-1)$ students are all distinct, because there are no two students who solved same two problems. On the other hand, besides student $A$ and these $3(k-1)$ students there are no others, because every student solved one of the problems solved by $A$. Therefore the number of the students is $n = 3(k-1) + 1 = 3k - 2$.

Let us now determine the number of the problems. Every student solved three problems, but every problem was solved by $k$ students. Therefore the number of the problems is $\frac{3n}{k}$ and it has to be a positive integer.

Using the formula above for $n$ we get
$$\frac{3n}{k} = \frac{3 \cdot (3k-2)}{k} = 9 - \frac{6}{k},$$
from which it follows that $k$ divides $6$, which implies $k \in \{2, 3, 6\}$ ($k \ge 2$).

For $k=2$ we have $n=4$ students $(A, B, C, D)$ and $\frac{3n}{k} = 6$ problems $(1, 2, 3, 4, 5, 6)$.
$A: 1, 2, 3 \quad B: 1, 4, 5 \quad C: 2, 4, 6 \quad D: 3, 5, 6$

For $k=3$ we have $n=7$ students and $\frac{3n}{k} = 7$ problems. For example, we can have:
$A: 1, 2, 3 \quad B: 1, 4, 5 \quad C: 1, 6, 7$
$D: 2, 4, 6 \quad E: 2, 5, 7 \quad F: 3, 4, 7 \quad G: 3, 5, 6$

For $k=6$ we would have $n=16$ students and $\frac{3n}{k} = 8$ problems. Then number of the pairs of the problems would be $\binom{8}{2} = 28$. However, number of the pairs of the problems solved by the same student is $3 \cdot 16 = 48$, because there are no two students who solved same two problems. Since $48 > 28$, this situation cannot happen.

Hence, the only possibilities are $(n, k) \in \{(1, 1), (4, 2), (7, 3)\}$.