This question resembles my previous question A cubic equation, and integers of the form $a^2+32b^2$ , but seems to be more difficult.

We are trying to determine whether there are any integers $x,y,z$ such that $$ 1 - 12 x^2 - 4 y^2 + y z^2 = 0. \quad\quad\quad (1) $$ It is clear that $z$ is odd. After multiplying by $-16$ and rearranging, we can rewrite the equation as $$ (8y-z^2)^2 + 192x^2 = z^4 + 16. $$ Writing $z=2t+1$ and denoting $s=8y-z^2$, we obtain $$ s^2+192x^2=P(t), \quad\quad\quad (2) $$ where $P(t)=(2t+1)^4+16$. If this equation has no integer solutions, then so is the original one.

To solve this, we need to understand what integers are representable as $s^2+192x^2$. This seems non-trivial even for primes. Cox's famous book "primes of the form $x^2+n y^2$" gives a characterization in terms of certain polynomial $P_n$, but contains no tables of $P_n$ for small $n$, so what is $P_{192}$ ?

In fact, I cannot even find online a list of reduced quadratic forms by discriminant long enough to contain $D=-4 \cdot 192 = -768$. In particular, what is the class number $h(-768)$?

But of course the main question is: are there any integers $x,y,z$ satisfying (1)?