The second four-solution conjecture for a parametrized exponential Diophantine equation

Let $m$ be a positive integer. Consider the Diophantine equation

$$x^2+\frac{1}{3}(2^{2m+6}-1)2^n=\frac{1}{9}\left(49\cdot 4^{2m+5}-11\cdot 4^{m+3}+1\right).$$

The second four-solution conjecture. This equation has exactly four solutions in integers $(x,n)$, with

$$n=0,\quad 3,\quad 2m+7,\quad 2m+8.$$

The conjecture is part of the discussion of equations of the form $x^2=A2^n+B$ with $A<0$; the supplied text gives no resolution status.