How can one show that the hyperelliptic curve $y^2 = x^{p} + \frac{1}{4}$ has only one positive rational solution for every prime $p>3$? Without applying Fermat's Last Theorem, how can one show that the hyperelliptic curve $y^2 = x^{p} + \frac{1}{4}$ has only one positive rational solution $(x,y) = (0, \frac{1}{2})$ for ever prime $p \geq 5$ ?

It is equivalent to FLT. Indeed, if $a^p+1=b^p$ for positive rational $a,b$, we have $x^p:=(ab)^p=a^p(a^p+1)=(a^p+1/2)^2-1/4:=y^2-1/4$.
Opposite implication (moving from the comments): if $x^p=y^2-1/4=(y-1/2)(y+1/2)$, $x\ne 0$, denote $y-1/2=a/m$ for coprime non-zero integers $a,m$. Then also $a+m\ne 0$, $y+1/2=(a+m)/m$ and $x^p=a(a+m)/m^2$. Since $a,a+m,m$ are mutually coprime and $p$ is odd, we get that they should be all perfect $p$-th powers, $a=A^p$, $m=B^p$, $a+m=C^p$, $A^p+B^p=C^p$.
