Substituting $z = x + y - 23$ from the first equation into the second yields $x^2 + y^2 - (x + y - 23)^2 = 23$ which simplifies to
$xy - 23x - 23y = -23 \cdot 12.$
Adding $23 \cdot 23$ to both sides and factoring yields
$$(x - 23)(y - 23) = 23 \cdot 11.$$
As $23$ and $11$ are primes, the only factors on the right hand side are $1$, $11$, $23$ and $11 \cdot 23$. Thus $x = 23 + 1 = 24$, $x = 23 + 11 = 34$, $x = 23 + 23 = 46$ or $x = 12 \cdot 23 = 276$; The corresponding values of $y$ are $276$, $46$, $34$, $24$ and the values of $z$ are $277$, $57$, $57$, $277$. The negative factors of $23 \cdot 11$ don't yield solutions, as $z$ would be negative.

Substituting $z = x + y - 23$ from the first equation into the second and simplifying like in Solution 1 yields
$$xy = 23x + 23y - 23 \cdot 12. \tag{1}$$
The right hand side of (1) is divisible by $23$, so the left side must be as well.
Thus $x$ or $y$ is divisible by $23$. Let $x = 23k$ (in the other case we may swap $x$ and $y$ due to symmetry). Substituting it into (1) and dividing the sides by $23$ yields $ky = 23k + y - 12$, from which $k = \frac{y-12}{y-23} = 1 + \frac{11}{y-23}$. For $k$ to be a positive integer, there are two possibilities:
* if $y - 23 = 11$, then $k - 1 = 1$, which yields $y = 34$, $x = 46$, $z = 57$;
* if $y - 23 = 1$, then $k - 1 = 11$, which yields $y = 24$, $x = 276$, $z = 277$.
In addition to those, we get the solutions with $x$ and $y$ swapped.