Find all $n>1$ odd number s.t. for any $x, y$ divisors of $n$ with $gcd(x,y)=1$ we have $x+y-1$ divide $n$.
It's intuitive that $n$ is of the form $p^k$ with $p$ prime.
My idea:
I consider $n=p_1^{a_1}\cdot p_2^{a_2}\cdot ...\cdot p_k^{a_k}$ with $p_1<p_2<...<p_k$.
$k\geq 2$
I let $x=p_1$ and $y=p_2^{a_2}\cdot...\cdot p_k^{a_k}$.
But $p_2,...,p_k$ don't divide $x+y-1$ because of minimality of $p_1$. So $x+y-1$ should be a power of $p_1$. But this is not necessary false and I am stuck.
Only prime powers can satisfy the desired hypothesis. As you rightly observed, if $p$ is the smallest prime factor of $n$, with $p^m\mid\mid n$, and $n$ is not a prime power, then with $x=p$ and $y=np^{-m}$ such that $(x+y-1)\mid n$, it follows that $x+y-1= p^l$ for some natural number $l\le m$; equivalently, $$n=(1-p+p^l)p^m\,.$$ Clearly, we must have that $l\ne 1$, thus in particular $m\ge 2$. We claim that we can find a different pair $(x,y)$ failing the desired hypothesis. This time, let $$x=p^2\,,~\,~\,~\,~\text{and}\,~\,~\,~\,~y= 1-p+p^l\,.$$ In particular, $n=xyp^{m-2}$. It is now not difficult to check that $(x+y-1)\nmid n$; indeed, a common prime factor of both $n$ and $x+y-1=-p+p^2+p^l$ must either divide $y-1$ and $x$ (which is simply $p$) or must divide $y$ and $x-1=p^2-1=(p+1)(p-1)$, which, because $p$ is odd, cannot exist because it would necessarily be smaller than $p$, contrary to the minimality of $p$. We therefore conclude that if $(x+y-1)|n$, then $p$ is the only prime factor of $x+y-1=p(-1+p+p^{l-1})$, which is blatantly absurd!
Remarks
The argument shows that if one allows for $n$ to be even, the last part would require that $p=2$ such that $(2^{l-1}+1)|(2^l-1)$. This is only possible if $l=2$, leading to the only potential even non-prime power cases satisfying your hypothesis as $n\in\{2^{m}\cdot 3:m\ge 2\}$. Of these, letting $x=2^m$ and $y=3$ shows that necessarily $m=2$, and thus $n=12$ is the only admissible solution.