Fermat-like Equations

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Problem 678

If a triple of positive integers $(a, b, c)$ satisfies $a^2+b^2=c^2$, it is called a Pythagorean triple. No triple $(a, b, c)$ satisfies $a^e+b^e=c^e$ when $e \ge 3$ (Fermat's Last Theorem). However, if the exponents of the left-hand side and right-hand side differ, this is not true. For example, $3^3+6^3=3^5$.

Let $a, b, c, e, f$ be all positive integers, $0 \lt a \lt b$, $e \ge 2$, $f \ge 3$ and $c^f \le N$. Let $F(N)$ be the number of $(a, b, c, e, f)$ such that $a^e+b^e=c^f$. You are given $F(10^3) = 7$, $F(10^5) = 53$ and $F(10^7) = 287$.

Find $F(10^{18})$.