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

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

It is of course possible to consider compositions of the above formulæ 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 $\{1,2,\dotsc\}^d$ and $\{1,2,\dotsc\}$ where $s_1,s_2,\dotsc$ 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 $\{1,2,\dotsc\}^d\longrightarrow \{1,2,\dotsc\}$)?

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