Hint $\rm\,\ m,n\mid x\!\iff\! mn\mid mx,nx\!\iff\! mn\mid\overbrace{(mx,nx)}^{\textstyle (m,n)x}\!$ $\rm\iff\! \overbrace{mn/(m,n)}^{\textstyle \ell}\mid x$
Let $\rm\:x = \ell\:$ in $\,(\Leftarrow)\,$ to get $\rm\:m,n\mid \ell,\:$ i.e. $\rm\:\ell\:$ is a common multiple of $\rm\:m,n,\:$ necessarily the least common multiple, since $(\Rightarrow)$ shows $\rm\:m,n\mid x\:\Rightarrow\:\ell\mid x\:\Rightarrow\:\ell\le x.$
Remark $\ $ That $\rm\: m,n\mid x\iff \ell\mid x\,$ is a definition of $\rm\,{\rm lcm}(m,n)\,$ in more general rings since - as above - it implies that $\,\ell\,$ is a common multiple of $\rm\,a,b\,$ that is divisibly least, i.e. it divides every common multiple. See this answer for this universal approach to LCMs and GCDs.
One easy and insightful way is to use the proof below. It essentially constructs $\rm\:gcd\:$ from $\rm\:lcm\:$ by employing duality between minimal and maximal elements - see the Remark below. This is essentially how the linked Wikipedia proof works, but there the innate duality is obfuscated by the presentation. Below is a proof structured so that this fundamental duality is brought to the fore.
$\rm{\bf Theorem}\quad c\mid a,b\iff c\mid d,\ \ $ for $\rm\ \ d = ab/lcm(a,b).\ $ $\rm\color{#0a0}{Hence}$ $\rm\ d = gcd(a,b)$
$\rm{\bf Proof}\qquad\ \ \, c\mid a,b \iff a,b\mid ab/c \iff lcm(a,b)\mid ab/c \iff c\mid ab/lcm(a,b)$
$\rm\color{#0a0}{Generally}\,$ if $\rm\, c\mid a,b\iff c\mid d\ $ then $\rm\ d = \gcd(a,b)\ $ up to unit factors, i.e. they're associate.
Indeed setting $\rm\:c = d\:$ in direction $(\Leftarrow)$ shows that $\rm\:d\mid a,b,\:$ i.e. $\rm\:d\:$ is a common divisor of $\rm\:a,b.\:$ Conversely, by direction $(\Rightarrow)$ we deduce that $\rm\:d\:$ is divisible by every common divisor $\rm\:c\:$ of $\rm\:a,b,\:$ thus $\rm\:c\mid d\:\Rightarrow\: c\le d,\:$ so $\rm\:d\:$ is a greatest common divisor (both divisibility and magnitude-wise).
Remark $\ $ The proof shows that, in any domain, if $\rm\:lcm(a,b)\:$ exists then $\rm\:gcd(a,b)\:$ exists and $\rm\ gcd(a,b)\,lcm(a,b) = ab\ $ up to unit factors, i.e. they are associate. The innate duality in the proof is clarified by employing the involution $\rm\ x'\! = ab/x\ $ on the divisors of $\rm\:ab.\:$ Let's rewrite the proof using this involution (reflection).
Notice that $\rm\ x\,\mid\, y\:\color{#c00}\iff\: y'\mid x'\,\ $ by $\smash[t]{\,\ \rm\dfrac{y}x = \dfrac{x'}{y'} \ }$ by $\rm\, \ yy' = ab = xx',\ $ so rewriting using this
$\begin{eqnarray}\rm the\ proof\ \ \ c\mid a,b &\iff&\rm b,\,a\mid ab/c &\iff&\rm lcm(b,\,a)\mid ab/c &\iff&\rm c\mid ab/lcm(b,a)\\[.5em]
\rm becomes\ \ \ \ c\mid a,b &\color{#c00}\iff&\rm a',b'\mid c' &\iff&\rm lcm(a',b')\mid c' &\color{#c00}\iff&\rm c\mid lcm(a',b')'\end{eqnarray}$
Now the innate duality is clear: $\rm\ gcd(a,b)\,=\,lcm(a',b')'\ $ by the $\rm\color{#0a0}{above}$ gcd characterization.
Best Answer
Forget about the condition $m\leq n$ for the moment. Since $600=2^3\cdot 3^1\cdot 5^2$ we have $$m=2^{\alpha_2}3^{\alpha_3}5^{\alpha_5},\quad n=2^{\beta_2}3^{\beta_3}5^{\beta_5}$$ with $\alpha_i$, $\beta_i\geq0$ and $$\max\{\alpha_2,\beta_2\}=3,\quad \max\{\alpha_3,\beta_3\}=1,\quad \max\{\alpha_5,\beta_5\}=2\ .$$ It follows that $$\eqalign{(\alpha_2,\beta_2)&\in\{(0,3),(1,3),(2,3),(3,3),(3,2),(3,1),(3,0)\}\>,\cr (\alpha_3,\beta_3)&\in\{(0,1),(1,1),(1,0)\}\>,\cr (\alpha_5,\beta_5)&\in\{(0,2),(1,2),(2,2),(2,1),(2,0)\}\cr}$$ are admissible, allowing for $7\cdot3\cdot5=105$ combinations. Exactly one of them has $m=n$, namely $m=n=600$, and in all other $104$ cases $m\ne n$. Since we want $m\leq n$ we have to throw out half of these cases, leaving $52+1=53$ different solutions of the problem.