Write $f_n = 2^{2^n} + 1$ and $a_n = a^{2^n} + 1$. Suppose now, if possible, that $f_n$ and $a_n$ are both prime for all but finitely many non-negative integers $n$, say, for all $n > N$. Clearly, we may and will assume that $f_n > a$ for all $n > N$.

Fix any $n > N$ and consider the (multiplicative) order of $a$ modulo the prime $f_n$. By Fermat's Little Theorem, this order divides $f_n - 1 = 2^{2^n}$, so it is a power of $2$, say, $2^k$.

Thus, the prime $f_n$ divides $a^{2^k} - 1 = (a^{2^{k-1}} - 1)a_{k-1}$, so it divides (at least) one of the two factors. By minimality of $k$, the first factor is not divisible by $f_n$, so $f_n$ divides $a_{k-1}$. Notice that $a_{k-1} \ne f_n$, since $a$ is not a power of $2$. Hence $a_{k-1}$ is composite, forcing $k-1 \le N$.

Consequently, the product $a_0 a_1 \dots a_N$ is divisible by $f_n$ for all $n > N$. This contradiction concludes the proof.