Find all $m,n \in \mathbb N$ Such that $3(2^m)+4=n^2$. 
Find All $n,m\in \mathbb N$ Such that $3(2^m)+4=n^2$

I’ve tried to plug some values of $m$, and it turns up that the only valid values is $$m\in \{2,5,6\}.$$
So once i saw that i tried to prove that it doesn’t exist a solution to the equation $\forall m\geq7$. But i have no idea how to prove this.
Something i’m not sure whether it’s true:
$$3(2^m)+4=n^2 \iff3(2^m+1)=n^2-1 \\ \iff 3(2^m+1)=(n+1)(n-1)$$
Thus $3\mid (n+1)(n-1) $, But since $3$ is a prime and $\gcd(n+1,n-1)=1$ or $2 \implies$
$3\mid n+1$ or $3\mid n-1$ What’s next?

In respect of the stated comment, I add our second case (and so the whole solution) I left for OP to solve.
First, check $m=1$.
If $m≥2, m-2=k$ and $n=2n_1$, then
$$3×2^{k}+1=n_1^2, n_1≥2$$
Case $-1.$
$n_1=3a+1$, where $a≥1.$
$$3×2^k=9a^2+6a$$
$$2^k=3a^2+2a$$
$$a≥1 \Longrightarrow \begin{cases} a=2b \\ k≥3 \end{cases}$$
$$2^k=12b^2+4b$$
$$2^{k-2}=3b^2+b=b(3b+1)$$
$$b=2^x, 3b+1=2^y$$
If $x≥1$, then we get $3b+1≥7$ and $3b+1$ is an odd number, which gives a contradiction.
So, we deduce that $x=0.$
Then, we can check $n_1=3a+2$ by the same way.
Of course, $n_1=3a$ is impossible.

The method of looking at our second case is the same as the first one, as I mentioned earlier.
Case $-2.$
$n_1=3a+2$, where $a≥0.$
First, check $a=0$. If, $a≥1$ then
$$3×2^k+1=9a^2+12a+4$$
$$2^k=3a^2+4a+1, a≥1$$
$$a≥1 \Longrightarrow \begin{cases} a=2b-1 \\ k≥3 \end{cases}$$
We have,
$$2^k=3(2b-1)^2+4(2b-1)+1, b≥1$$
$$2^k=12b^2-4b$$
$$2^{k-2}=3b^2-b=b(3b-1)$$
$$b=2^x, 3b-1=2^y$$
If $x≥1$, then we get $3b-1≥5$ and $3b-1$ is an odd number, which gives a contradiction.
So, we deduce that $x=0.$
Finally, backward all the steps and you will find all required integer values ​​of $m$ and $n.$
