Show that the polynomial $xyz + x + z$ takes on all sufficiently large integer values when $x \geq 2$ and $y,z \geq 1$ are integers.
We show that the only positive integers that cannot be expressed as $xyz+x+z$ with $x\geq2$ and $y,z\geq1$ integers are: $1,2,3,4,6,18$.
First of all, it is clear that $1,2,3,4$ cannot be expressed as $xyz+x+z\geq5$ when $x\geq2$ and $y,z\geq1$. Moreover, $xyz+x+z\geq7$ when $x\geq3$ and $y,z\geq1$, hence if $6$ could be expressed as $xyz+x+z$, we would have $x=2$ and thus $6=(2y+1)z+2$, i.e., $2y+1$, an odd number greater than $2$, would divide $4$, which is not the case. In turn, $6$ cannot be expressed as $xyz+x+z$ within the given domain ($x,y,z$ positive integers with $x\geq2$).
Similarly, if one had $18=xyz+x+z$ with $x\geq2$ and $y,z$ all positive integers, then $xy+1$ would divide $18-x$ as $18-x=z(xy+1)$. However, it is pretty clear that $16$ has no factor of the form $2y+1$, $15$ has no factor of the form $3y+1$, $14$ has no factor of the form $4y+1$, $13$ has no factor of the form $5y+1$, $12$ has no factor of the form $6y+1$. For $x\geq7$, the only integer $xy+1$ that has a chance to divide $18-x$ is $x+1$ since $2x+1>18-x$. Furthermore, $2(x+1)>18-x$, hence the only divisibility possibility is that $x+1=18-x$, which also proves impossible since $18$ is even. All in all, $18$ cannot be expressed $xyz+x+z$ with $x\geq2$ and $y,z$ all positive integers.
Let us now prove that all the other positive integers can be represented as stated. With $x=2$, the formula is conveniently factored as $2+(2y+1)z$. As a result, any integer $n\geq3$ such that $n-2$ has one odd factor can be expressed in the given form. Consider thus numbers of the form $2^k+2$ with $k$ positive integer different than $0,1,2,4$ (which correspond to the numbers $3,4,6,18$ respectively). We can distinguish multiple cases based on $k$.
If $k$ is odd, then $k\geq3$ since $k\neq1$, and $2^k+2 \equiv 1[3]$ (since $2^2 \equiv 1[3]$ and $2^1+2 \equiv 1[3]$). As a result, $2^k+2 = 3p+1$ for some $p\geq3$ and thus
$$ 2^k+2 = 3p+1 = p\times2\times1 + p + 1 = xyz+x+z $$
with $x=p\geq2$ and $y=2,z=1$.
If $k$ is even but not divisible by $4$, then $k\geq6$ (since $k\neq2$) and $2^n+2\equiv1[5]$ (since $2^4\equiv1[5]$ and $2^2+2\equiv1[5]$). Hence, $2^k+2 = 5p+1$ for some $p\geq13$ and thus
$$ 2^k+2 = 5p+1 = p\times4\times1 + p + 1 = xyz+x+z $$
with $x=p\geq2$ and $y=4,z=1$.
If $k$ is divisible by $4$, there are three further subcases to distinguish. We write $k=4m$ with $m\geq2$ (as $k\neq4$) and distinguish based on $m$.
If $m$ is divisible by $3$, then $m\geq3$ and $2^{4m}+2 \equiv 3[7]$ (as $2^{12}\equiv1[7]$ and $2^0+2\equiv3[7]$). We may thus write $2^k+2 = 7p+3$ for some integer $p\geq585$, and therefore
$$ 2^k+2 = 7p+3 = p\times2\times3 + p + 3 = xyz+x+z
$$
with $x=p\geq2$ and $y=2,z=3$.