<s>[INST] 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?
 [/INST] (Here's a suggested approach. If you're stuck, explain where you're stuck and what you've tried.)

*

*Factor it as $ 3 \times 2^m = (n-2) ( n+2)$.

*Hence $ \{ n-2, n+2 \} = \{ 3 \times 2^a, 2^b \}$.

*Hence $  3 \times 2^a = 2^b \pm 4 $.

*If $ b = 0, 1, 2$, what can we say? What are the solutions with these small cases?


 $3 \times 2^a = 1+4, 2+4, 4+4$ (cannot take negative branch).
 So we get $ b=1, a = 1 \Rightarrow 2 \times 6 = 3\times 2^2  $.


*

*If $ b \geq 3$, what can we say?


 RHS is a multiple of 4 but not 8, so $ a = 2$.
$3 \times 4 = 2^b \pm 4$, which gives us $b=3, 4$.
$a = 2, b = 3 \Rightarrow 8\times 12 = 3\times 2^5 .$
$a = 2, b = 4 \Rightarrow 12 \times 16 = 3\times 2^6 . $

In conclusion, we have solutions only when $ m = 2, 5, 6$.
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