<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] First of all, let's complete the square on the RHS and write it as $(3n-2)^2+15$.  Next, suppose $m$ is odd.  Then $2^mm^2$ will be $\equiv 2\pmod 4$ (can you see why?) but the RHS — as a sum of a square and a number $\equiv 3\pmod 4$ — will be congruent to either $3$ or $0$, so they can't be equivalent.
Now, suppose $m$ is even; then $m=2r$, say, and so the LHS $m^22^m = (2r)^22^{2r} = \left(r2^{r+1}\right)^2$.  This means we can write $a^2=b^2+15$, where $a=r2^{r+1}$ and $b=3n-2$.  This can be written perhaps more suggestively as $a^2-b^2=15$; can you take it from there?
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