Let $a$ and $b$ be positive integers.
If $b$ is even (i.e. $b=2c$ for some positive integer $c$), then $a^2 - 2^b$ can be factored as $$a^2 - 2^b = a^2 - 2^{2c} = (a + 2^c)(a - 2^c).$$
Edited: (October 2, 2021 - 4:18 PM Manila time) I would have to qualify that I am only interested in those $a$ such that $a$ is odd.
Here is my question:
What about if $b$ is odd (i.e. $b=2d+1$ for some nonnegative integer $d$)? Can $a^2 - 2^b$ still be factored?
MY ATTEMPT
Let $X = a^2 - 2^b$, and let $b=2d+1$ for some nonnegative integer $d$.
Then we obtain $$X = a^2 - 2^{2d+1}$$ $$X - 2^{2d+1} = a^2 - 2^{2d+2} = (a + 2^{d+1})(a - 2^{d+1})$$ from which we finally get the expression $$X = 2^{2d+1} + (a + 2^{d+1})(a - 2^{d+1}) = 2^b + (a + 2^{(b+1)/2})(a - 2^{(b+1)/2})$$ for $X$.
Alas, this is where I get stuck!
If we allow irrational coefficients, then it can be factored: we have $$a^2-2^{2d+1}=a^2-(\sqrt2\cdot2^d)^2=(a-\sqrt2\cdot2^d)(a+\sqrt2\cdot2^d).$$ However, if we are a bit stricter, then it cannot be factored in terms of $a$ and $2^d$. Suppose that there are polynomials $P(a,2^d)$ and $Q(a,2^d)$ with integer coefficients such that $P(a,2^d)Q(a,2^d)=a^2-2^{2d+1}$. Then it follows that $P(x,y)Q(x,y)=x^2-2y^2$ for all $x,y$. But since $x^2-2y^2$ is irreducible over $\mathbb Z[x,y]$, the only options are the trivial ones where one of $P$ and $Q$ is the constant polynomial $1$.