Perhaps there are none with integral coefficients; so let us admit rational coefficients. The map $(x, y) \mapsto x + \frac{1}{2}(x + y)(x + y + 1)$ is well known, and swapping $x$ and $y$ in the formula yields another, so we have two for starters.
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3$\begingroup$ There have been a number of questions related to this, including one of the highest-voted ones by Bjorn Poonen. You might search through existing questions. $\endgroup$– Will JagyCommented Nov 7, 2010 at 15:59
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4$\begingroup$ See mathoverflow.net/questions/9731/… and mathoverflow.net/questions/21003/… $\endgroup$– Will JagyCommented Nov 7, 2010 at 16:03
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1$\begingroup$ In particular, the first link tells us that this question is an open problem. $\endgroup$– Martin BrandenburgCommented Nov 7, 2010 at 16:15
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$\begingroup$ @Martin actually that is only asking about surjectivity when the domain is $\mathbb{Z}\times\mathbb{Z}$, but I agree that it has some bearing here. $\endgroup$– David Roberts ♦Commented Nov 8, 2010 at 2:49
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Describing such bijections is an open problem. Maximal result (there is no other bijections among polynomials of degree not higher than 4) are contained in
John S. Lew, Arnold L. Rosenberg, Polynomial indexing of integer lattice-points I. General concepts and quadratic polynomials, J. Number Theory 10 (1978) pp 192-214, doi:10.1016/0022-314X(78)90035-5.
Polynomial indexing of integer lattice-points II. Nonexistence results for higher-degree polynomials, J. Number Theory 10 (1978) pp 215-243, doi:10.1016/0022-314X(78)90036-7
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$\begingroup$ What, exactly, are you claiming to be an open problem? As far as I can see, the question asks for a list of examples. $\endgroup$ Commented Mar 12, 2016 at 9:45
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$\begingroup$ Ok, Describing such bijections is an open problem. $\endgroup$ Commented Mar 12, 2016 at 16:30