There are the nice and simple identities,
$$(2x+1)^2 + (2x^2+2x-1)^2 - (2x^2+2x)^2 = 2$$
$$ (6x^3+1)^3 + (-6x^3+1)^3 + (-6x^2)^3 = 2$$
$$ (8x^5-2x)^4 + (8x^4+1)^4 + (8x^4-1)^4 - (8x^5+2x)^4 - (4x^2)^4 = 2$$
$$ (-8x^6+2x)^5 + (-8x^6-2x)^5 + (8x^5+1)^5 + (-8x^5+1)^5 + 2(8x^6)^5 = 2$$
with the first two presumably discovered by multiple authors, while $k=4$ is by Seiji Tomita and $k=5$ by Ajai Choudhry.
Q: Can we find the $k=6$ or $k=7$ version?
P.S. Jarek Wrobleski found,
$$(1 + 3^5\sqrt3\, x^7)^7 + (1 - 3^5\sqrt3\, x^7)^7 + (3x - 3^6x^8)^7 + (-3x - 3^6x^8)^7 + 2(3^6x^8)^7 = 2$$
though since this is a Diophantine problem and to be consistent with the rest, terms should be integers.