$\begin{align}{\bf Hint}\ \
&\gcd(X,Y) = d,\ \ \ {\rm lcm}(X,Y) = m \ \ \ \text{yields by $\rm\color{#90f}{cancelling}$ $\,d\neq 0$}\\[.3em]
\iff\ &\color{#0a0}{gcd(x,\,y)\ \ = 1},\qquad\ \ \, \color{#c00}{x\cdot y}\ =\, m/d,\ \ {\rm for}\ \ x = X/d,\,\ y = Y/d
\end{align}$
$\begin{align}{\rm e.g.}\ \ \
&\gcd(X,Y) = 15,\ \, {\rm lcm}(X,Y) = 150\\[.3em]
\iff\ &\gcd(x,\,y)\ \ =\ 1,\qquad\ \ \, x\cdot y\,\ =\,\ 10
\end{align}$
So it's equivalent to factoring $\,m/d = \color{#c00}{xy}\,$ into ${\rm \color{#0a0}{coprime}\ factors}\,\ \color{#0a0}{x,y}$.
This method quickly and simply solves all your problems.
$\rm\color{#90f}{Cancelling}$ to reduce to a coprime case is a common way to simplify homogeneous divisibility problems.
Using notation $\begin{align}[x,y] &:= {\rm lcm}(x,y),\\ (x,y)&:= \gcd(x,y)\end{align}\,$ and $\,\color{#c00}{\rm D}:=$ lcm distributes over gcd
$\qquad\underbrace{[q,n]=(\color{#0af}{[q,m]},[q,n])}_{\textstyle \color{#0af}{[q,m]}=[q,n]} \overset{\color{#c00}{\rm D}}= [q,\color{#0a0}{(m,n)}]\ {\large \mid}\ [q,\color{#0a0}{m\!+\!n}]\ $ since $\ \color{#0a0}{(m,n)\mid m\!+\!n}$
Best Answer
This is an updated edit to my previous answer, which was misguided in assuming facts not presented in the question as posed.
The numbers $6,4,21$ appear in the question. $6=2\cdot 3, 4=2^2, 21=3\cdot 7$, so to answer the question it will be necessary to keep track of the prime factors $2,3,7$ as they may appear in $m$ and $n$. In what follows, all variable names are natural numbers.
Let $m=(2^a3^b7^c\cdot d)$ and $n=(2^e3^f7^g\cdot h), \gcd(d,h)=1$. Since there is one factor of each of $2$ and $3$ in $m$ and $n$, $\min(a,e)=1, \min(b,f)=1$; that is, $2$ and $3$ must appear as factors in each of $m,n$ at least once, but in one or the other of them only once. Since $m$ and $n$ have no common factor of $7$, $\min(c,g)=0$; that is, at least one of $m,n$ must have no factor of $7$.
Next, $4m=2^{a+2}3^b7^c\cdot d$ and $21n=2^e3^{f+1}7^{g+1}\cdot h$. Let $i=\min(a+2,e), j=\min(b,f+1), k=\min(c,g+1)$. Then lcm$4m\cdot 21n=\frac{84mn}{2^i3^j7^k}$.
$i$ may take on the values $1,2,3$; $j$ may take on the values $1,2$; $k$ may take on the values $0,1$. so there are twelve possible values for the denominator $2^i3^j7^k$ (which are $6,12,18,24,36,42,72,84,126,168,252,504$) and hence twelve possible answers to the posed question. A single answer would require knowing more about the factors of $m$ and $n$, which information is not given.