Integer points on a surface I would like to understand ${\bf nonnegative \  integral}$ solutions $(x,y,z)$ on the surface 
$$xyz-ax-ay-bz=d.$$
where $a,b,d$ are positive integers.

I can certainly prove that for a fixed $z$ there are finitely many non-negative solutions. But I'm having difficult time proving that it has finitely many solutions, even for some special triples $a,b,d$.
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.



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 $y\ge x$.
There will always be finitely many (positive) solutions. 

Supposing $a, b, c, d$ are all positive integers, then the positive integer solutions to 

$$xyz = ax + by + cz + d \tag{1}$$

are bounded, for the following reason:

Observe that $(1)$ is equivalent to

$$\begin{align} z &amp; = {ax + by + d \over xy + c} \\ &amp; \le {ax + by + d \over xy} \tag2 \\ &amp;  = \frac ay + \frac bx + \frac d{xy} \\ &amp; \le a + b + d \tag3 \end{align}$$

$(2)$ because the new denominator is smaller (since $c \gt 0$), and $(3)$ similarly because $x, y \ge 1$.

By an identical argument you will find $1 \le x \le b + c + d$ and $1 \le y \le a + c + d$

Let $M = 3 \cdot \max \{a, b, c, d\}$ just for simplicity's sake; then all positive integer solutions to $(1)$ are bounded within the region $[1, M] \times [1, M] \times [1, M]$, within which there are only finitely many ($M^3$) integral points