Find m and n in the given equation: Sorry new to this forum and don't know how to format:

If $m,n\in\Bbb N$ satisfy $6^{2m+2}\cdot 3^n=4^n\cdot 9^{m+3}$, then $n$ and $m$ must be ... what?

This is for my discrete mathematics class. I tried taking ln/log of both sides but got nowhere. Primes didn't go anywhere for me either.
Edit: Maybe I have to use Fundamental Theorem of Arithmetic? 
 A: Hint $\rm\displaystyle\ \ \frac{\color{#c00}{3^n}}{\color{#0a0}{4^n}} = \frac{9^{m+3}}{(3\cdot 2)^{2(m+1)}} = \frac{9^{m+3}}{9^{m+1} 4^{m+1}} = \frac{\color{#c00}{9^2}}{\color{#0a0}{4^{m+1}}}\ \Rightarrow\ \color{#c00}{n = \ldots}\ \Rightarrow\ \color{#0a0}{m = n-1 = \ldots}$
Remark $\ $ This implicitly uses the uniqueness of reduced fractions, which is equivalent to uniqueness of prime factorizations (used in the other answers/hints).
A: Hint:
$$\begin{align*}6^{2m+2}\cdot 3^n&=4^n\cdot 9^{m+3}\\ \\
\longrightarrow \color{red}{2^{2m+2}}\cdot\color{blue}{3^{2m+2+n}}&=\color{red}{2^{2n}}\cdot \color{blue}{3^{2m+6}}
\end{align*}$$
Two integers are the same iff they have the same prime factors. Thus you can equate the exponents corresponding to each prime and solve for the resulting system of equations.
A: We have the following:
$$
6^{2m+2}\cdot 3^n=(2\cdot 3)^{2m+2}\cdot 3^n = 2^{2m+2}\cdot 3^{2m+2}\cdot 3^n = 
2^{2m+2}\cdot 3^{2m+2+n}
$$
And
$$
4^n\cdot 9^{m+3} = 2^{2n} \cdot 3^{2m+6}.
$$
Equating them gives:
$2^{2m+2}\cdot 3^{2m+2+n} = 2^{2n} \cdot 3^{2m+6}.$. So we need to find $m,n \in \mathbb{N}$ such that $2m+2=2n$ and $2m+2+n=2m+6$, because prime factorization is unique.
The rest I will leave up to you.
