You do not need the final $d^3.$ Every integer is the sum of two squares and a cube, as long as we do not restrict the $\pm$ sign on the cube.
TYPESET FOR LEGIBILITY:
Solution by Andrew Adler:
$$ 2x+1 = (x^3 - 3 x^2 + x)^2 +(x^2 - x - 1)^2 -(x^2 - 2x)^3 $$
$$ 4x+2 = (2x^3 - 2 x^2 - x)^2 +(2x^3 -4x^2 - x + 1)^2 -(2x^2 - 2x-1)^3 $$
$$ 8x+4 = (x^3 + x +2 )^2 +(x^2 - 2x - 1)^2 -(x^2 + 1)^3 $$
$$ 16x+8 = (2x^3 - 8 x^2 +4 x +2)^2 +(2x^3 -4x^2 - 2 )^2 -(2x^2 - 4x)^3 $$
$$ 16x = (x^3 +7 x - 2)^2 +(x^2 +2 x + 11)^2 -(x^2 +5)^3 $$
You can check these with your own computer algebra system. Please let me know if I mistyped anything.
Alright, our conjecture (Kaplansky and I) is that, for any odd prime $q,$ $x^2 + y^2 + z^q$ is universal. However, this is false as soon as the exponent on $z$ is odd but composite. The example we put in the article is
$$ x^2 + y^2 + z^9 \neq 216 p^3, $$
where $p \equiv 1 \pmod 4$ is a (positive) prime.
This defeated a well-known conjecture of Vaughan. We told him about it in time for him to include it in the second edition of his HARDY-LITTLEWOOD BOOK, where it is now mentioned on pages 127 ("There are some exceptions to this,") and exercise 5 on page 146.