Do there exist $m,n$ such that $6 = 2(2m+1)^2/(2n+1)^2$? Can the two numbers $n$ and $m$ exists such that
$$6=\frac{2(2m+1)^2}{(2n+1)^2}$$
where $\gcd(m,n) = 1$?

So you want to prove that without using the fact that $\sqrt{3}$ is irrational.
No problem.
Suppose that you have $3=\dfrac {(2m+1)^2}{(2n+1)^2}$.
Then $(2m+1)^2=3(2n+1)^2$.
That is, $4m^2+4m+1=12n^2+12n+3$.
So $4m^2+4m-12n^2-12n=2$ so the left hand side is divisible by $4$ but the right hand side is not, so such $m,n$ do not exist.
