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The question is to find all integer solutions to the equation $$ x^2+y^2=z^3+1. $$ This equation obviously has infinitely many integer solutions (take, for example, $(x,y,z)=(1,u^3,u^2)$ for any integer $u$) but the question is to describe all integer solutions. A parametric representation (like this answer to Solve for integers $x, y$ and $z$: $x^2 + y^2 = z^3.$ for equation $x^2+y^2=z^3$) would of course be ideal, but, in general, by "describe" I mean any algorithm that explicitly produces all solutions without performing any search by trial and error. See my previous question Solve in integers: $y(x^2+1)=z^2+1$ for an example of such algorithm. After that equation has been solved, the equation in question is one of the smallest/simplest ones for which I am not aware about any reasonable method/algorithm to describe all solutions, hence the question.

Update 04.02.2022: After the equations $y^2+z^2 = 2x^2 \pm 1 $ has been solved by individ in the comment to this question, we can now explicitly describe the solution sets to all equations of size $h \leq 17$, where $h$ is defined here What is the smallest unsolved diophantine equation? , except for the following equations of size $17$:

$$ y^2+z^2 = x^3+1 $$ $$ y^2+z^2 = x^3-1 $$ $$ y(x^2-y)=z^2-1 $$

All these equations have infinitely many integer solutions, but the question is to find all solutions. If you know any explicit way to describe all solutions to any of these equations, please answer.

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    $\begingroup$ You linked to an MSE question, but appeared to intend to link to the answer; I edited accordingly. I would also encourage you to word your questions as questions or requests, rather than as imperatives (e.g., "How does one solve this equation?" rather than "Solve this equation"). $\endgroup$
    – LSpice
    Commented Feb 1, 2022 at 21:30
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    $\begingroup$ Ok, title reformulated as a question, as suggested. $\endgroup$
    – Bogdan Grechuk
    Commented Feb 1, 2022 at 21:52
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    $\begingroup$ If a prime $p$ divides $z^3+1$, then $z=kp-1$ for some integer $k$. Conversely, if $z=kp-1$ with $p\ne3$ prime, then $z^3+1=kp(k^2p^2-3kp+3)$ is a multiple of $p$ but not of $p^2$. So the question of whether there's a solution is very closely related to the question of whether $z+1$ is a multiple of some prime $p\equiv3\bmod4$. $\endgroup$
    – Gerry Myerson
    Commented Feb 1, 2022 at 22:58
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    $\begingroup$ Hm, a prime $7$ divides $3^3+1=28$, but $3$ is not of the form $7k-1$ for some integer $k$. $\endgroup$
    – Bogdan Grechuk
    Commented Feb 1, 2022 at 23:07
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    $\begingroup$ What Bogdan said — the fact that $z^3+1$ only has the real root $z=-1$ over $\mathbb{R}$ doesn't mean that it can't have three roots in $\mathbb{Z}/p\mathbb{Z}$ for some $p$; it depends on whether $-3$ is a quadratic residue mod $p$ or not,. $\endgroup$
    – Steven Stadnicki
    Commented Feb 1, 2022 at 23:15

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