Solution:
Answer: $(1, 1)$.

It is clear that for a given integer $n$, there is at most one integer $k$ for which the equation holds. For $n=1$ this is $k=1$. But, for $n>1$, problem 1 asserts that $\sigma(n) \phi(n) \leq n^{2}-1 < n^{2}$, so that $k \geq 2$. We now claim that $2 > \frac{n^{2}}{\sigma(n) \phi(n)}$. Write $n = p_{1}^{e_{1}} \cdots p_{k}^{e_{k}}$, where the $p_{i}$ are distinct primes and $e_{i} \geq 1$ for all $i$, and let $q_{1} < q_{2} < \cdots$ be the primes in ascending order. Then
$$\begin{aligned} & \frac{n^{2}}{\sigma(n) \phi(n)} = \prod_{i=1}^{k} \frac{p_{i}^{2 e_{i}}}{\frac{p_{i}^{e_{i}+1}-1}{p_{i}-1} \cdot (p_{i}-1) p_{i}^{e_{i}-1}} = \prod_{i=1}^{k} \frac{p_{i}^{2 e_{i}}}{p_{i}^{2 e_{i}} - p_{i}^{e_{i}-1}} \\ & \quad = \prod_{i=1}^{k} \frac{1}{1 - p_{i}^{-1-e_{i}}} \leq \prod_{i=1}^{k} \frac{1}{1 - p_{i}^{-2}} < \prod_{i=1}^{\infty} \frac{1}{1 - q_{i}^{-2}} \\ & \quad = \prod_{i=1}^{\infty} \left( \sum_{j=0}^{\infty} \frac{1}{q_{i}^{2j}} \right ) = \sum_{n=1}^{\infty} \frac{1}{n^{2}} \\ & \quad < 1 + \sum_{n=2}^{\infty} \frac{1}{n^{2}-1} = 1 + \frac{1}{2} \left( \left( \frac{1}{1} - \frac{1}{3} \right ) + \left( \frac{1}{2} - \frac{1}{4} \right ) + \left( \frac{1}{3} - \frac{1}{5} \right ) + \cdots \right ) = \frac{7}{4} < 2. \end{aligned}$$
It follows that there can be no solutions to $k = \frac{n^{2}}{\sigma(n) \phi(n)}$ other than $n = k = 1$.