The answer is that $m$ and $n$ should be powers of the same prime number. These all work because for a prime power we have
$$\frac{\sigma(p^e) - 1}{p^e - 1} = \frac{(1 + p + \cdots + p^e) - 1}{p^e - 1} = \frac{p(1 + \cdots + p^{e-1})}{p^e - 1} = \frac{p}{p-1}.$$
So we now prove these are the only ones. Let $\lambda$ be the common value of the three fractions.

Claim — Any solution $(m, n)$ should satisfy $d(mn) = d(m) + d(n) - 1$.

*Proof.* The divisors of $mn$ include the divisors of $m$, plus $m$ times the divisors of $n$ (counting $m$ only once). Let $\lambda$ be the common value; then this gives
$$\begin{aligned} \sigma(mn) &\ge \sigma(m) + m\sigma(n) - m \\ &= (\lambda m - \lambda + 1) + m(\lambda n - \lambda + 1) - m \\ &= \lambda mn - \lambda + 1 \end{aligned}$$
and so equality holds. Thus these are all the divisors of $mn$, for a count of $d(m) + d(n) - 1$. $\square$

Claim — If $d(mn) = d(m) + d(n) - 1$ and $\min(m, n) \ge 2$, then $m$ and $n$ are powers of the same prime.

*Proof.* Let $A$ denote the set of divisors of $m$ and $B$ denote the set of divisors of $n$. Then $|A \cdot B| = |A| + |B| - 1$ and $\min(|A|, |B|) > 1$, so $|A|$ and $|B|$ are geometric progressions with the same ratio. It follows that $m$ and $n$ are powers of the same prime. $\square$