Since you say you cannot use the simple min/max exponents proof via unique factorization, here is a proof that uses only universal gcd laws (so will work in any gcd domain). We simply eliminate all lcms by $\rm\:[x,y] = xy/(x,y),\:$ and apply gcd laws (distributive, commutative, associative, etc).
$$\rm\begin{eqnarray}
\rm &\rm\qquad\qquad (a,[b,c])\ &=&\rm\ [(a,b),(a,c)] \\
\rm \iff&\rm\qquad\quad \left(a,\dfrac{bc}{(b,c)}\right)\ & =&\rm\ \dfrac{(a,b)(a,c)}{(a,b,c)} \\
\iff &\rm (a,b,c)(a(b,c),bc)\ &=&\rm\ (a,b)(a,c)(b,c)
\end{eqnarray}$$
which is true since both sides $\rm = (aab,aac,abb,abc,acc,bbc,bcc)\:$ by distributivity etc.
If you are not proficient with gcd laws, you may find it helpful to rewrite the proof employing a more suggestive arithmetical notation, namely denoting the gcd $\rm (a,b)\:$ by $\rm\ a \dot+ b.\:$ Because the arithmetic of GCDs shares many of the same basic laws of the arithmetic of integers, the proof becomes much more intuitive using a notation highlighting this common arithmetical structure. Below is a sample calculation comparing the two notations.
$$\rm\begin{eqnarray}
\rm(a,\:b)\ (a,\:c) &=&\rm (a(a,\!\:c),b(a,\!\:c)) &=&\rm ((aa,ac),\:(ba,bc)) &=&\rm (aa,ac,\:\!ba,\:bc) \\
\rm\ (a\dot+ b)(a\dot+c) &=&\rm \color{#c00}{a(a\dot+c)}\dot+b(a\dot+c) &=&\rm (\color{#c00}{aa\dot+ac})\dot+(ba\dot+bc) &=&\rm aa\dot+ac\dot+ba\dot+bc
\end{eqnarray}$$
Now the gcd arithmetic looks like integer arithmetic, e.g. uses of the gcd distributive law look the same as for integers, e.g. $\ \rm \color{#c00}{a(a\!+\!c) = aa\!+\!ac}\,$ etc.
Best Answer
Hint $ $ For $\,x = \gcd(m,n),\ {\rm lcm}(m,n) = mn/x,\,$ and your equation is $\,(x-m)(x-n) = 0.$ Thus either $\,x = \gcd(m,n) = m,\,$ so $\,m\mid n,\ $ or $\,x = \gcd(m,n) = n,\,$ so $\,n\mid m.$