Consider the map $$f \colon \mathbb N^2 \to \mathbb N$$ defined as $f(n,m)=2^n(2m +1) - 1$ this is primitive recursive and it's also bijective.
Bijectivity follows directly by the unique factorization's theorem: indeed for every natural number $x \in \mathbb N$ we have that $x+1 > 0$ can be uniquely factor as the product of finite sequence power of primes $x=2^{n}p_1^{n_1}\dots p_k^{n_k}$ where the $p_i$ are odd primes, the product $p_1^{n_1}\dots p_k^{n_k}$ is odd so it can be expressed as $2m + 1$ for a unique $m \in \mathbb N$. So $x+1 = 2^n(2m+1)$ for a unique pair $(n,m) \in \mathbb N^2$ and so $x= 2^n(2m + 1) - 1$.
For the primitive recursion part, we have that the functions
$$g \colon \mathbb N \to \mathbb N$$
$$n \mapsto 2^n$$
$$h \colon \mathbb N \to \mathbb N$$
$$m \mapsto 2m + 1$$
the multiplication
$$ \cdot \colon \mathbb N^2 \to \mathbb N$$
and the predecessor
$$p \colon \mathbb N\setminus \{0\} \to\mathbb N$$
$$p(n) = n-1 $$
are primitive recursive, so the function $f=p \circ (\cdot)\circ (g,h)$ which is obtained by composition by these function is primitive recursive too.