How does the fact that Fermat primes are relatively prime imply there are infinite 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).