Are there any nonzero rational solutions to this equation? Are there any nonzero rational solutions to the equation 
$$y^2 = 64x^n + 1$$ where $n\geq 3$ is an integer ?
For the case $n=3$, the question can be settled by basic ideas of elliptic curves, but i'm not sure of the higher cases ?
 A: When $n=3$, the group of rational solutions is given by the points $(x,y)\in \{(0,\pm 1), (1/2,\pm 3),(-1/4,0)\}$, along with the point at infinity.  The only $x$ value among these points which is a (proper) perfect power is $0$, so the only solutions when $3|n$ and $n>3$ are the trivial ones.
When $n=4$, we'll assume joro's Sage computation, that there are no non-trivial rational points.  This also deals with the case when $4|n$.
Now, we may as well replace $n$ by its largest prime factor, so we reduce to dealing with the case when $n=p\geq 5$ is prime.  We can quickly check that $y\neq 0$, and reduce to looking for solutions when $y>0$.  Write $y=a/b$ with $a,b$ positive, relatively prime integers.  Similarly, write $x=c/d$ with $c\in \mathbb{Z}-\{0\}$, $d>0$ an integer, and $\gcd(c,d)=1$.  We then have
$$
a^2-b^2=2^6b^2\left(\frac{c}{d}\right)^p.
$$
The LHS is an integer, hence so is the RHS.  We will look at two main cases.
Case 1: $b$ is odd.
For any prime $q|b$, the LHS has $q$-adic valuation $0$, hence so does the RHS.  So all of the primes in $b$ are accounted for in $d$.  There are thus two cases, depending on whether $2$ divides $d$:
Case 1a: Say $d$ is odd.  Then $d^p=b^2$.  Thus, $b=s^p$, $d=s^2$, for some odd integer $s>0$.  Then we have
$$
(a-s^p)(a+s^p)=2^6c^p.
$$
The two factors on the left hand side have GCD at most $2$.  On the other hand, at least one of the factors must in fact be divisible by $2$.  Hence either
$$
a-s^p=2u^p,\qquad a+s^p=2^5v^p
$$
or
$$
a-s^p=2^5u^p,\qquad a+s^p=2v^p
$$
where $c=uv$ with $\gcd(u,v)=1$.  Solving for $a$, and plugging into the other, and dividing by $2$, we get
$$
(\ast)\qquad s^p+u^p=2^4v^p \qquad \text{ or } \qquad s^p + 2^4u^p = v^p.
$$
Note that these are really the same equation, after moving all terms to one side, using the fact that we can bring $-1$ into a $p$th power.
Case 1b: $d$ is even.  Then (by similar reasoning as in case 1a) $p=5$, $b=s^5$, $d=2s^2$, for some odd integer $s>0$.  Then we have
$$
(a-s^5)(a+s^5)=2c^5.
$$
Now, $a$ must be odd, so that $2$ divides the LHS.  But then the LHS has at least $2$-adic valuation $\geq 2$, hence $2|c$, which contradicts $\gcd(c,d)=1$.
Case 2: $b$ is even.  Say $b=2^{k}e$ for some odd integer $e>0$, and some integer $k\geq 1$.
By the same reasoning as in case 1, we must have that $d^p=2^{2k+6}e^2$, hence (since $p$ is odd) we get $k=pt-3$ for some integer $t\geq 1$, $d=2^{2t}s^2$, and $e=s^p$ for some odd integer $s>0$.  Hence, we obtain
$$
(a-2^{pt-3}s^p)(a+2^{pt-3}s^p)=c^p
$$
with $c$ odd.  The factors on the LHS are relatively prime, and equal to a $p$-power, hence are both $p$-powers.  Thus, solving as before, we eventually arrive at
$$
(\ast\ast)\qquad \frac{1}{4}w^p + u^p=v^p
$$
where $w=2^ts$.

In other words, moving terms accordingly, you can reduce your equation to solving the two Fermat-like equations
$$
X^p+Y^p=16Z^p\qquad \text{ or } \qquad 4X^p+4Y^p=Z^p.
$$
I'm not certain if the modularity methods for solving Fermat's last theorem apply here, but that's probably the next place to look.
