Consider Diophantine equation:
$$(m-4)n=2m.$$
It can be rewritten as:
$$\frac{1}{n}+\frac{2}{m}=\frac{1}{2}.$$
If we search for positive solutions only, we have at least $4$ pairs:
$$(m,n)\in\{(5,10);(6,6);(8,4);(12;3)\}$$
They can be illustrated in the following way using regular polygons:
How are the equation and the polygon combinations related? Do we have any other positive solutions?
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Its $n=\frac{2m}{m-4}$ and thus $m-4 \mid 2m$. It follows that $m-4 \mid 2m-2(m-4)=8$. So $m-4=1, 2, 4, 8$ and $m=5, 6, 8, 12$.
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In a regular polygon with $m$ sides, each internal angle measures $$ 180°\cdot{m-2\over m}. $$ If you join two such equal polygons along a common side, they form an outside angle of $$ 360°- 2\cdot180°\cdot{m-2\over m}=360°\cdot{2\over m}. $$ You can go on joining other polygons, always on the same side: the chain will close if the angle computed above is the internal angle of a regular polygon with $n$ sides, for some value of $n$. That happens if: $$ 360°\cdot{2\over m}=180°\cdot{n-2\over n}, $$ which is equivalent to your equation. The solutions of this equation then represent the only possible "regular polygon closed chains" one can obtain. See the other answers for a proof that there are only four possible solutions.
It's $$mn-2m-4n=0$$ or $$(m-4)(n-2)=8$$ and we get a finite number of cases:
$m-4=1$ and $n-2=8$, which gives $(m,n)=(5,10)$;
$m-4=2$ and $n-2=4$, which gives $(m,n)=(6,6)$;
$m-4=4$ and $n-2=2$, which gives $(m,n)=(8,4)$ and
$m-4=8$ and $n-2=1$, which gives $(m,n)=(12,3)$.
Easy to see that these all natural solutions.