Consider two integers $x$ and $y$ and their respective prime factorizations \begin{equation} x = p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}} \dots p_{n}^{\alpha_{n}} \end{equation} and \begin{equation} y = q_{1}^{\beta_{1}}q_{2}^{\beta_{2}} \dots q_{m}^{\beta_{m}} \end{equation}
If all of the prime factors above are distinct (i.e. $p_i \neq q_j$ for all $i \neq j$ in $\{1, 2, \dots, \max(m, n)\}$), then can we say that $x$ and $y$ are co-prime?
I feel like the answer is yes, since (edit: the following is incorrect) the only factors of $x$ are the $p_1, p_2, \dots, p_n$ and similarly the only factors of $y$ are the $q_1, q_2, \dots, q_m$. But, I'm not sure how this changes when we're considering powers and products of these factors.
Suppose $x,y$ are not coprime. Then they have a common divisor $d$. If $d\mid x$, then the prime factors of $d$ must be selected from $p_i$. Similarly, if $d\mid y$, then the prime factors of $d$ must be selected from $q_j$. But $p_i \ne q_j$, so $d$ has no prime factors; $d=1$ and $x,y$ are coprime.