here is my proof attempt:
$\Rightarrow$ Direction;
Since $(a,b) = 1$, $\exists x,y \in \mathbb{Z}$ $\ni$
$ax + by = 1$
But $p \vert a$ $\Rightarrow$ $p \vert ax$
Suppose towards a contradiction that $p \vert b$, then $p \vert by$, hence
$p \vert ax + by$
$p \vert 1$
Contradiction since $p$ is prime, $\therefore$ $p \nmid b$.
$\Leftarrow$ Direction;
Let $a = p_1^{e_1} p_2^{e_2} \cdot \cdot \cdot p_m^{e_m}$ and $b = q_1^{f_1}q_2^{f_2} \cdot \cdot \cdot q_n^{f_n}$ ; not all $e_i,f_i$ zero.
We want to show
$p_i^{e_i} \neq q_j^{e_j}$ (**) for no $i,j$ between 1 and $m,n$ resp.
But $p \vert a$ $\Rightarrow$ $p = p_i^{e_i}$ some $i, 1 \leq i \leq m$
By our assumption, $p \vert a$ $\Rightarrow$ $p \nmid b$
$\Rightarrow$ $p = p_i^{e_i} \neq q_j^{f_j}$ for any $j$.
Hence (**) satisfied, ($a$, $b$ share no common prime factors), $\Rightarrow$ $(a,b) = 1$
I use the FTOA when rewriting $a$ and $b$.