The so-called Cantor bijection given by 
$$(x,y)\longmapsto {x+y\choose 2}-{x\choose 1}+1$$
is a bijection from $\{1,2,3,\ldots\}^2$ onto $\{1,2,3,\ldots\}$.

It can be generalized to bijections $\{1,2,3,\ldots\}^d
\longrightarrow \{1,2,3,\ldots\}$ by considering 
$$(x_1,\ldots,x_d)\longmapsto (d+1\mod 2)+(-1)^d\sum_{k=1}^d(-1)^k{x_1+\ldots+x_k\choose k}$$
where $(d+1\pmod 2)$ equals $1$ if $d$ is even and $0$ otherwise. (The proof is a sort of double induction on $d$ and on the sum $x_1+x_2+\ldots+x_d$.)

It is of course possible to consider compositions of the above formulae in order to get additional, more complicated polynomial bijections.

A straightforward counting argument shows that we obtain in this way $d! s_d$ different polynomial bijections between $\{1,2,\ldots\}^d$ and $\{1,2,\ldots\}$ where $s_1,s_2,\ldots$
are the Schroeder numbers with generating series
$$\sum_{n=1}^\infty s_nq^n=\frac{1+q-\sqrt{1-6q+q^2}}{4}\ .$$

Are there other "exotic" polynomial bijections (between $\{1,2,\ldots\}^d\longrightarrow \{1,2,\ldots\}$)?

(The answer is obviously "no" for $d=1$ and unknown for $d=2$.
I ignore if an "exotic" bijection is known for $d=3$.)