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.
Best Answer
Denote the number of admissible orderings of $n$ dwarfs by $a_n$. Let $n\geq2$. If $\gamma:\>[n]\to[n]$ is an admissible ordering then $$\bar\gamma(k):=n+1-\gamma(k)\qquad(1\leq k\leq n)$$ is an admissible ordering as well. The map $\gamma\mapsto\bar\gamma$ is in fact a fixed point free involution of the set of admissible orderings. Among $\gamma$, $\bar\gamma$ exactly one ends with a downward slope of the zigzag, and exactly one starts with an upward slope of the zigzag.
Assume there are $n+1$ dwarfs. The largest among them (the boss) can range himself at $n+1$ different places in the row. For a given choice there are $0\leq k\leq n$ free places ahead of the boss and $n-k$ free places behind the boss. We can choose the $k$ dwarfs marching in front of the boss in ${n\choose k}$ ways. If $k=0$ or $k=1$ we can arrange these $k$ dwarfs in one way, and if $k\geq2$ we can arrange them in ${1\over2}a_k$ ways such that their zigzag ends with a downward slope. Similarly, if $k=n-1$ or $k=n$ we can arrange the remaining $n-k$ dwarfs to the right of the boss in one way, and if $k\leq n-2$ we can arrange these dwarfs in ${1\over2}a_{n-k}$ ways such that their zigzag begins with an upward slope.
We therefore obtain the following recursion: $$a_0=a_1=1,\quad a_2=2,\quad a_3=4\ ,$$ $$a_{n+1}={1\over2}a_n+{n\over2}a_{n-1}+{1\over4}\sum_{k=2}^{n-2}{n\choose k}a_ka_{n-k}+{n\over2}a_{n-1}+{1\over2}a_n\quad(n\geq3)\ .$$ This recursion produces the following table: $$\bigl(a_n\bigr)_{n\geq0}=\bigl(1, 1, 2, 4, 10, 32, 122, 544, 2770, 15872, 101042,\ldots\bigr)\ .$$ This is sequence A001250 at OEIS, where they refer to the greek word $\beta\omicron\upsilon\sigma\tau\rho\omicron\phi\eta\delta\omicron\nu$, meaning going back and forth.