Proving that if $n$ is odd and $\gcd(m, n) = 1$, then $\gcd(2m + n, 2n) = 1$

Since we have that $(m,n)=1$, there are $x,y$ so that $$ mx+ny=1\tag{1} $$ Then, $$ \begin{align} (\color{#C00000}{2m+n})2x+\color{#C00000}{2n}(2y-x)&=\color{#00A000}{4}\\ (\color{#C00000}{2m+n})n-\color{#C00000}{2n}m&=\color{#00A000}{n^2} \end{align}\tag{2} $$ Since $n$ is odd, there are $w,z$ so that $$ 2w+nz=1\tag{3} $$ Squaring $(3)$ gives $$ \color{#00A000}{4}(w^2+wnz)+\color{#00A000}{n^2}z^2=\color{#0000FF}{1}\tag{4} $$ Combining $(2)$ and $(4)$, we get an integral combination of $\color{#C00000}{2m+n}$ and $\color{#C00000}{2n}$ that equals $\color{#0000FF}{1}$: $$ (\color{#C00000}{2m+n})\left(2x(w^2+wnz)+nz^2\right)+\color{#C00000}{2n}\left((2y-x)(w^2+wnz)-mz^2\right)=\color{#0000FF}{1}\tag{5} $$ Therefore, $(\color{#C00000}{2m+n},\color{#C00000}{2n})=\color{#0000FF}{1}$.

It can easily be deduced that $d(=gcd(2m+n,2n))$ will be odd because if it is not then $2|(2m+n) \Rightarrow 2|n$ which is not true.
Now ($d|2n \Rightarrow d|n$ and $d|(2m+n) \Rightarrow d|m$) $\Rightarrow d|gcd(m,n)\Rightarrow d|1\Rightarrow d=1$