I think you are trying to look only at how $n$ is expressed in terms of the rest of the variables. Indeed, such an approach is good, but you can do with other variables as well! So let's try to write $n$ in terms of $y$, by eliminating $x$ from both equations.
From the first, $x = ny-3n-3$. From the second, $x = \frac{y+n}{3} + n$.
Combining these, we get $\frac{y+n}{3} + n = ny-3n-3$,so multiplying by $3$, $y+4n = 3ny-9n-9$. Isolating $n$ gives $y + 9 = n(3y-13)$, and hence $$
n = \frac{y+9}{3y-13}
$$
Note that $n$ is a positive integer. If $y > 12$, then $3y-13 > 22$ while $y + 9 < 22$, so that $n$ cannot possibly be an integer! Furthermore, $3y-13 < 0$ if $y < 5$, so we have $5 \leq y \leq 11$.
Now, we can test : $y = 5$ gives $n = 7$ and $x = 11$, which works out.
$y = 6$ gives $n = 3$ and $x = 6$, which works out.
$y = 7$ gives $n = 2$ and $x = 5$, which works out.
$y = 8,9,10$ don't work out. Finally, $y = 11$ gives $n = 1$ and $x = 5$ which works out.
Finally, $n = 1,2,3,7$.
Concluding, the difference of approach was in isolating a different variable from that of $n$. Remember that all the variables are related in a manner such that knowing one means you know the rest, so it was sufficient to deal with any of the variables. Finally, expressions involving $\frac{ay+b}{cy+d}$ are restrictive if integers, when $a,c$ are small.
Step 1: Obtain $a+c=2$.
Step 2:
Note that
$$ab+bc+cd+ad=(a+c)(b+d)$$
Adding four equations gives
$$(a+c)(b+d)+2(a+c)+2(b+d)=16$$
$$(a+c)(b+d+2)+2(b+d+2)=20$$
$$(a+c+2)(b+d+2)=20$$
With $a+c=2$, we have $b+d=3$.
Step 3:
Further manipulating, 1-2+3-4 gives
$$(a-c)(b+d)=6$$
Thus, $a-c=2$.
Therefore, we have $a=2, c=0$.
Step 4: Put $a,c$ into 1,
$$2b+d=3$$
With $b+d=3$, $b=0, d=3$.
Best Answer
Yes, we can can get a sum of squares here. We need to prove that $$(x^2+y^2+z^2)^2\geq3xyz(x+y+z)$$ or $$\sum_{cyc}(x^4+2x^2y^2-3x^2yz)\geq0$$ or $$\sum_{cyc}(2x^4-2x^2y^2+6x^2y^2-6x^2yz)\geq0$$ or $$\sum_{cyc}(x^4-2x^2y^2+y^4+3(x^2z^2-2z^2xy+y^2z^2))\geq0$$ or $$\sum_{cyc}((x^2-y^2)^2+3z^2(x-y)^2)\geq0.$$