nice problem! ^_^ Firstly note that $p$ divides neither $a$ or $b$ -- if $p$ divides one of them, $p$ must divide the other, contradicting the fact that $\gcd(a, b) = 1$. Thus, we can write that $$a^{2^n} + b^{2^n}\equiv 0\pmod{p}\iff \left(\frac{a}{b}\right)^{2^n} + 1 \equiv 0\pmod{p}.$$Furthermore, we can obtain that $$\left(\frac{a}{b}\right)^{2^n} \equiv -1\pmod{p} \iff \left(\frac{a}{b}\right)^{2^{n+1}}\equiv 1\pmod{p},$$and so $\mathrm{ord}_p\left(\frac{a}{b}\right) \mid 2^{n+1}.$ However, since $$\left(\frac{a}{b}\right)^{2^n}\equiv -1\pmod{p},$$we must have that $\mathrm{ord}_p\left(\frac{a}{b}\right) = 2^{n+1},$ as if the order is $2^d$ for some $0\leq d\leq n$, then we would have $$\left(\frac{a}{b}\right)^{2^n}\equiv 1\pmod{p},$$a contradiction. To finish, note that by FLT, $$\left(\frac{a}{b}\right)^{p-1} \equiv 1\pmod{p},$$and so we must have that $2^{n+1} = \mathrm{ord}_p\left(\frac{a}{b}\right) \mid p-1,$ as desired.