The proof is by induction. The base is provided by the $n=0$ case, where $7^{7^0} + 1 = 7^1 + 1 = 2^3$.

To prove the inductive step, it suffices to show that if $x = 7^{2^{m-1}}$ for some positive integer $m$ then $(x^7+1)/(x+1)$ is composite. As a consequence, $x^7+1$ has at least two more prime factors than does $x+1$.

To confirm that $(x^7+1)/(x+1)$ is composite, observe that
$$\begin{aligned} \frac{x^7+1}{x+1} &= \frac{(x+1)^7 - ((x+1)^7 - (x^7+1))}{x+1} \\ &= (x+1)^6 - \frac{7x(x^5 + 3x^4 + 5x^3 + 5x^2 + 3x + 1)}{x+1} \\ &= (x+1)^6 - 7x(x^4 + 2x^3 + 3x^2 + 2x + 1) \\ &= (x+1)^6 - 7^{2m}(x^2 + x + 1)^2 \\ &= \left[(x+1)^3 - 7^m(x^2 + x + 1)\right]\left[(x+1)^3 + 7^m(x^2 + x + 1)\right] \end{aligned}$$
Also each factor exceeds 1. It suffices to check the smaller one; $\sqrt{7x} \le x$ gives
$$\begin{aligned} (x+1)^3 - 7^m(x^2+x+1) &= (x+1)^3 - \sqrt{7x}(x^2+x+1) \\ &\geq x^3 + 3x^2 + 3x + 1 - x(x^2+x+1) \\ &= 2x^2 + 2x + 1 \geq 113 > 1. \end{aligned}$$
Hence $(x^7 + 1)/(x + 1)$ is composite and the proof is complete.