Funny problem. Let $P$ be the product of all prime numbers which divides $m$, but doesn't divide $n$. Therefore, we have $P | m$, and $\gcd(P,n) = 1$. Set $x = P, y = n$, we have \[ \gcd(P + n, mn) > 1 \]Since $\gcd(P + n, n) = \gcd(P,n) = 1$, we then have \[ \gcd(P + n, m) > 1 \]Suppose there exists a prime $p$ dividing $m$ and $P + n$. Case 01. If $p$ divides $n$ as well, then $p$ doesn't divide $P$ by definition, a contradiction. Case 02. If $p$ doesn't divide $n$, this means $p | P$ by definition, which is again a contradiction.

Not much different from above but will post it any way. Let $p_1,p_2,\dots ,p_k$ be all primes that divide both $m,n$ and let $q_1,q_2,\dots ,q_l$ be set of primes that divide $m$ but not $n$ and let $r_1,r_2,\dots ,r_t$ be the set of primes that divide $n$ but not $m$ it is easy to check that $x=p_1 p_2 \dots p_kq_1q_2 \dots q_l,y=r_1r_2 \dots r_t$ has the property that $\gcd(x+y,mn)=1$.