Integer points on a surface

If $x=0$, then $-ay-bz=d$ where LHS is non-positive while RHS is positive. So, we have $x\not=0$. We have $y\not=0$ and $z\not=0$ similarly.

Now, we may suppose that $y\ge x\gt 0$ and $z\gt 0$.

If $\min(x,y,z)=z$, then $$1=\frac{a}{yz}+\frac{a}{xz}+\frac{b}{xy}+\frac{d}{xyz}\le \frac{a}{z^2}+\frac{a}{z^2}+\frac{b}{z^2}+\frac{d}{z^3}$$ $$\implies z^3-(2a+b)z-d\le 0$$ The number of $z$ satisfying this inequality is finite.

Consindering $$(zx-a)(zy-a)=bz^2+dz+a^2$$ for each $z$, we see that the number of $(x,y)$ is finite.

If $\min(x,y,z)=x$, then, similarly as above, the number of solutions is finite.

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Example :

For your example where $a=9,b=4,d=37$, if $\min(x,y,z)=z$, we have $$z^3-22z-37\le 0\implies z\le 5$$

For each $z=1,2,\cdots, 5$, consider $$(zx-9)(zy-9)=4z^2+37z+81$$

For example, for $z=1$, we have $$(x-9)(y-9)=2\times 61$$ $$\implies (x-9,y-9)=(1,2\times 61),(2,61)$$ $$\implies (x,y)=(10,131),(11,70)$$ since we already supposed that $y\ge x$.