<s>[INST] Solve integer equation $2^m.m^2=9n^2-12n+19$ 
Problem: Find $m,n\in \mathbb{N}^*$ satisfies: $2^m.m^2=9n^2-12n+19$.

This is my attempt:
We have $9n^2-12n+19\equiv 1 \pmod 3$, so: $2^{m}m^2\equiv 1 \pmod 3\tag{1}$
In addition, we have: $$m^2\equiv 0\text{ or }1 \pmod 3 $$
So $(1)\implies m\equiv \pm 1\pmod 3$.
Suppose that: $m=3k+1(k\in \mathbb{N})$.
We have: $$2^{m}m^2=2^{3k+1}(3k+1)^2=8^{k}(3k+1)^22\equiv 2^{k+1}(3k+1)^{2}\pmod 3\\ \implies k\equiv 1\pmod 2$$
I can only come here!!
 [/INST] The case left open by the above answer is for $m$ odd. If $m$ is odd, then $m^2\times 2^m = 2A^2$ for some integer $A$; namely $A = m\times 2^{\frac{m-1}{2}}$. Thus we are left with the equation
$$2A^2 = (3n-2)^2+15$$
Note that the LHS of the equation is either 2 or 0 mod 3, whereas the RHS is always 1.
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