I was just reading a book called Proofs from the Book. It presented the proof given by George Polya to prove that two Fermat primes (numbers of the form $2^{2^n} + 1$) are always relatively prime, which in itself is a very elegant proof. But, to say that it implies there are infinitely many primes does not make sense to me.
15 and 16 are relatively prime but it doesn't imply there are infinitely many primes.
Any sequence $F_n$ of pairwise coprime integers will contain infinitely many primes, because each new $F_n$ has a new prime factor, for all $n\in \mathbb{N}$. In your example, $n=15,16$, but we have infinitely many $n$.
Let $M_n=2^n-1$ be the $n$th Mersenne number, and $F_n=2^{2^n}+1$ be the $n$th Fermat number.
$$M_{2^{n}}=2^{2^{n}}-1=\prod_{k=0}^{n-1}(2^{2^{k}}+1)=\prod_{k=0}^{n-1}F_{k}.\tag{1}$$ Hence $$ F_{n}=\frac{M_{2^{n+1}}}{M_{2^{n}}}=\frac{\sum_{k=0}^{2^{n+1}-1}2^k}{\sum_{k=0}^{2^{n}-1}2^k} =\frac{\sum_{k=0}^{M_{n+1}}2^k}{\sum_{k=0}^{M_{n}}2^k}=\frac{2^{2^{n+1}}-1}{2^{2^{n}}-1}=2^{2^{n}}+1.\tag{2}$$
As $2$ is prime, Mersenne numbers $M_{2^k}$, with a subscript a power of $2$ only inherit factors from those preceding Mersenne numbers $M_{2^{k-j}}$ subscripts being a lesser power of $2$, $1\leqslant j\leqslant k-1$. These inherited factors are seen to be the Fermat numbers using (1):
\begin{align*} M_{2}&=F_{0}=(2^{2^{0}}+1)\\ M_{4}&=F_{0}F_{1}=(2^{2^{0}}+1)(2^{2^{1}}+1)\\ M_{8}&=F_{0}F_{1}F_{2}=(2^{2^{0}}+1)(2^{2^{1}}+1)(2^{2^{2}}+1)\tag{3}\\ M_{16}&=F_{0}F_{1}F_{2}F_{3}=(2^{2^{0}}+1)(2^{2^{1}}+1)(2^{2^{2}}+1)(2^{2^{3}}+1)\\ M_{32}&=F_{0}F_{1}F_{2}F_{3}F_{4}=(2^{2^{0}}+1)(2^{2^{1}}+1)(2^{2^{2}}+1)(2^{2^{3}}+1)(2^{2^{4}}+1) \end{align*} As such the $M_{2^n}$s primitive prime factors (meaning prime factors not appearing as factors before in the sequence of Mersenne numbers) are the factors of the $(n-1)$th Fermat number, the primitive prime factors in bold being factors of Fermat numbers: \begin{align*} M_{2}&=\mathbf{3}\\ M_{4}&=3\cdot\mathbf{5}\\ M_{8}&=3\cdot5\cdot\mathbf{17}\\ M_{16}&=3\cdot5\cdot17\cdot\mathbf{257}\tag{4}\\ M_{32}&=3\cdot5\cdot17\cdot257\cdot\mathbf{65537}\\ M_{64}&=3\cdot5\cdot17\cdot257\cdot\mathbf{641}\cdot65537\cdot\mathbf{6700417} \end{align*}
Theorem: Fermat numbers are pairwise coprime.
Proof: As $2$ is prime $M_{2^k}\mid M_{2^{k+1}}$ and $M_{2^{k+1}}/M_{2^{k}}$ is an integer being a product of $M_{2^{k+1}}$s primitive factors, hence
$$\gcd(F_{k},F_{k+1}) =\gcd\left(\frac{M_{2^{k+1}}}{M_{2^{k}}},\frac{M_{2^{k+2}}}{M_{2^{k+1}}}\right)=1,$$
is the greatest common divisor of adjacent Fermat numbers is that of the primitive prime factors of $M_{2^{k+1}}$ and $M_{2^{k+2}}$ which is necessarily $1$. It follows the Fermat numbers will be pairwise relatively prime as for any $k$, $j$ $\in\Bbb{N}$,
$$\gcd(F_{k},F_{j})=\gcd\left(\frac{M_{2^{k+1}}}{M_{2^{k}}},\frac{M_{2^{j+1}}}{M_{2^{j}}}\right)=1,$$
where $\gcd(F_{k},F_{j})$ is the greatest common divisor of the primitive prime factors of $M_{2^{k+1}}$ and $M_{2^{j+1}}$.
Q.E.D.
As the Fermat numbers are pairwise coprime this proves the infinitude of primes as each $F_n$ is a product of distinct primes, these being the primitive factors of the Mersenne numbers $M_{2^{n+1}}$, as illustrated by (2), (3), and (4).
To show that there are infinitely many prime numbers, it suffices to show that ever Fermat number is relatively prime to all the previous Fermat numbers. Then we can show that the prime decomposition of each Fermat number has none of the same prime factors as the prime decompositions of all the previous Fermat numbers. It can be shown that each Fermat number is relatively prime to all the previous Fermat numbers by showing that each Fermat number is the product of all the previous Fermat numbers plus 2. In can be proven by induction that each Fermat number is the product of all the previous Fermat numbers as follows. 5 = 3 + 2. 17 = (3 × 5) + 2. 257 = (15 × 17) + 2 = (3 × 5 × 17) + 2. 65537 = (255 × 257) + 2 = (3 × 5 × 17 × 257) + 2.
