There exists $t\in\mathbb{F}_{p^2}$ such that $t+1/t\equiv s_0\equiv 1/4 \pmod p$. Further, $s_i\equiv t^{2^i}+t^{-2^i} \pmod p$. As $p-1=\frac{2^n-2}3$, we can find remainder of $s_n$ bu division on $p$, that is $$s_n\equiv t^{2^n}+t^{-2^n}\equiv t^2+t^{-2}\equiv s_1 \pmod p.$$After that we can conclude that at most $4$ possible options for $s_{n-2}$ are valid (one of them is $3/2$), but here I get stuck.

Since $p\equiv3\pmod8$ and $n\not\equiv0\pmod3$, we get $p\equiv1,4\pmod7$, so $\left(\frac{-2}p\right)=\left(\frac{14}p\right)=1$. Let $s_n\equiv x^{2^{n+2}}+x^{-2^{n+2}}\pmod p$. where $x+x^{-1}\equiv\sqrt{\frac72}\pmod p$. Then, $x=\frac{\sqrt{14}+\sqrt{-2}}4$. We have \begin{align*} s_{n-2}&=x^{2^n}+x^{-2^n}\\ &=x^{3p-1}+x^{-3p+1}\\ &=x^{-1}x^{3p}+xx^{-3p}\\ &=\frac{\sqrt{14}-\sqrt{-2}}4\left(\frac{\sqrt{14}+\sqrt{-2}}4\right)^3+\frac{\sqrt{14}+\sqrt{-2}}4\left(\frac{\sqrt{14}-\sqrt{-2}}4\right)^3\\ &=x^2+x^{-2}\\ &=\frac32. \end{align*}

$n$ is odd and $p \equiv 3 \pmod{4}$. $s_0 = \frac{p+1}{4}$ from substitution. Using pavel's idea of $t + t^{-1} \equiv 4^{-1} \pmod{p}$, multiplying both sides by $t$ and $t^{-1}$ gives $t^2 - ts_0 + 1 \equiv 0 \pmod{p}$ and $t^{-2} - t^{-1}s_0 + 1 \equiv 0 \pmod{p}.$ By induction, $s_i \equiv t^{2^i} + t^{-2^i} \pmod{p}$ Since $2^n-2 = 3(p-1) \equiv 0 \pmod{\phi(p)}$, then $2^{n-1} \equiv 1 \pmod{p-1}$ so $s_{n-1} \equiv t + t^{-1} \equiv s_0 \pmod{p}$. Finally $s_{n-2}^2 \equiv s_{n-1}+2 \implies 2s_{n-2} \equiv \sqrt{4s_0 + 8} \equiv \sqrt{9} \pmod{p}$, and we're done.

At the first glance; this problme remided me an old idea applied https://artofproblemsolving.com/community/c6h15464p109517/

DottedCaculator wrote: where $x+x^{-1}\equiv\sqrt{\frac72}\pmod p$. Then, $x=\frac{\sqrt{14}+\sqrt{-2}}4$. That's nice idea to make two steps back!

aaravdodhia wrote: Since $2^n-2 = 3(p-1) \equiv 0 \pmod{\phi(p)}$, then $2^{n-1} \equiv 1 \pmod{p-1}$ That's not true, we can only say that $2^{n-1} \equiv 1 \pmod {\frac{p-1}2}$

$2^{n}+1=3p$ . If $n$ is even then $2^{n}+1 \equiv 2 (mod3)$, contradiction. So $n$ is odd. Then we have $-2$ is a square number of modulo $p$. So $p$ must be have the form of $8k+1$ or $8k+3$. On the other hand, $n>3$ so$2^n+1 \equiv 1 (mod 8) $. Therefore, $p = 8k+3$. Due to $n>3$ , $s_{0}=\dfrac{p+1}{4} >2 $ so there exist $a$ such that: $a+\dfrac{1}{a}=\dfrac{p+1}{4}$. By induction, we have: $s_{i}=a^{2^{i}}+a^{-2^{i}}$. $2^n+1=3p$ so $2^{n} - 2$ is divisible by $p-1$. $s_{n}=a^{2^{n}}+a^{-2^{n}} \equiv a^{2}+a^{-2}=s_{1}$ $ s_{n-1} \equiv \sqrt{s_{1}+2} (mod p) \Rightarrow s_{n-2} \equiv \sqrt{s_{0}+2}(modp)$ Therefore: $2s_{n-2} \equiv 3,-3 (modp)$. If $2s_{n-2} \equiv -3 (modp)$. Then $2s_{n-3}^2 \equiv 1 (modp)$ or $2$ is the squre number modulo $p$. So $p$ must be $8k+1$ (contradiction). Q.E.D