Substituting $y = -2^{x-1}$ into the original equation gives
$$f(0) = \frac{1}{2^{2^{x-1}}} f(f(x)) f(-2^{x-1}). \quad (1)$$
So, if $f(-2^x) = 0$ for at least one $x$ then also $f(0) = 0$. Then taking $x = 0$ and arbitrary $y$ in the original identity gives $f(1+2y) = 0$, i.e., $f \equiv 0$.

Assume in the rest that $f(-2^x) \neq 0$ for every $x$. Substituting $y = -2^x$ into the original equation gives $f(-2^x) = \frac{1}{2^{2^x}} f(f(x)) f(-2^x)$. Hence, for every $x$,
$$f(f(x)) = 2^{2^x} \quad (2)$$
Substituting (2) into the original equation and taking $y = 0$, we obtain
$f(2^x) = f(f(x))f(0) = 2^{2^x}f(0)$ for all $x$, which implies
$$f(x) = 2^x f(0) \quad (3)$$
for all positive $x$. On the other hand, applying (2) to (1) gives
$$f(-2^{x-1}) = \frac{2^{2^{x-1}}}{2^{2^x}} \cdot f(0) = 2^{-2^{x-1}} f(0)$$
for all $x$, which implies (3) also for all negative $x$.

We have shown above that $f(0) = 0$ implies $f(x) = 0$ for all $x$. Hence we may assume that $f(0)$ is either positive or negative. By taking $x = 0$ in (2) and applying (3), we obtain $2 = 2^{2^0} = f(f(0)) = 2^{f(0)} \cdot f(0)$. Both $f(0) < 1$ and $f(0) > 1$ would lead to contradiction, hence $f(0) = 1$ and the only non-zero solution is thus $f(x) = 2^x$.