I think you mean $1, d_1 + 1, d_2 + 1, . . . , d_k + 1$ are all the proper divisors of $m$

no, that was the official statement, as written

we can assume $1=d_1<d_2< \cdots<d_{k-1}<d_k=n$, $d_id_{k+1-i}=n$, $1, d_1 + 1, d_2 + 1, . . . , d_k + 1$ are all the proper divisors of $m$, then $m=n+1$, $2(d_{k-1}+1)=n+1$, $d_{k-1}=\frac{n-1}{2}$, $\frac{n-1}{2} \mid n$, then $n-1 \mid 2$, $n=2,3$. Oh! No! Are you sure this is original?

the official solution takes cases whether k=1 and whether not

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Any solution yet? Bump

I'm a bit late to the party but here's a quick sketch of a solution First take a look at the case when k=1. Then obviously n is the square of a prime so m must be a square of a prime too. We easily conclude that n=4 and m=9 are the only solutions. If k>1 then we sort the d-s such that d1<d2<...<dk. We see that d1dk=d2dk-1. Then (d1+1)(dk+1)=(d2+1)(dk-1+1) so when we multiply this we get that d1+dk=d2+dk-1. I haven't learnt latex yet so l can't write down the rest but it's simple casework and we get that the only other solution is (8,15)