Neat Diophantine Equation Question

Question

After some fairly tedious work including studying multiple different cases separately, I have found all the solutions to $$a^n+1=b^2 $$ where $a$, $b$, $n$ can take on the value of any integer, be it positive, negative or zero. I just wanted to see how MSE users would solve it perhaps there are easier and more elegant approaches.

(Note: I consider using Catalan's Conjecture at any stage of the argument to be "cheating". )

Answer 1

HINT:

$$a^n=b^2-1=(b-1)(b+1)$$

$(b+1,b-1)=(b+1,2)$

Case $\#1:$ If $a$ is odd, $b\pm1$ are odd, $(b+1,b-1)=1$

Case $\#2:$

If $a$ is even, $b\pm1$ are even $\implies n\ge2$ and $$\frac{a^n}4=\frac{b-1}2\cdot\frac{b+1}2$$

and $$\left(\frac{b-1}2,\frac{b+1}2\right)=1$$

Answer 2

if $a=2,n=3,b=3$ is only solution. Indeed if $n=2k+1$ we use Zsigmondy's theorem $$a^n+1=b^2$$ have no solution see Zsigmondy's theorem