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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 $$\begin{array}{rl}\sigma (mn)& \ge \sigma (m)+m\sigma (n)-m\\ & =(\lambda m-\lambda +1)+m(\lambda n-\lambda +1)-m\\ & =\lambda mn-\lambda +1\end{array}$$ 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. $\square$