How do I prove that the recurrence contains no perfect square? Given the recurrence
$$a_{n+2} = 14a_{n+1} - a_n - 6$$
with $a_1=1$ and $a_2=8$, how do I prove that none of the $a_n$'s apart from $a_1$ is a perfect square. This is not a homework problem, rather part of a small side project.
PS: I am aware that we can obtain a closed form solution for the recurrence, however that doesn't seem to help me answer the question, i.e., it is equivalent to proving that $\dfrac{(2+\sqrt3)^{2k-1}-(2-\sqrt3)^{2k-1}}{4\sqrt3} + \dfrac12$ is not a perfect square except for $k=1$.
EDIT: Elementary quadratic residue based arguments for small primes enable me to obtain that $n \equiv a \pmod b$, but doesn't seem to help me in ruling out all possible cases except $n=1$.
EDIT: To give the problem more context, I was trying to prove that the Mordell's equation $y^2=x^3+1$ has the solutions only as $(-1,0)$, $(0,\pm1)$ and $(2,\pm3)$. After some algebra, you need to find integer solutions to $2m^3-3m^2+1 = n^2$, which in turn results in the above claim.

The recursion preserves the quadratic form
$$ Q(x_n, x_{n+1}) = {x_{{n}}}^{2}-14\,x_{{n+1}}x_{{n}}+{x_{{n+1}}}^{2}+6\,x_{{n}}+6\,x_{{n
+1}}
$$
In this case $Q(1,8) = 7$.  Now the solutions of $Q(s,t) = 7$ are
$t = 7s - 3 \pm 4 \sqrt{3s^2 - 3s + 1}$.  If $s$ and $t$ are integers,
$3s^2 - 3 s + 1$ must be a square.  That is, $3 a_n^2 - 3 a_n + 1$ is a square.  With $a_n = x^2$, we are led to consider the curve
$$ 3 x^4 - 3 x^2 + 1 - y^2 = 0$$ 
which Maple says has genus $1$, so it's elliptic.  
Well, are there integer points other than $[0,\pm 1]$ and $[\pm 1, \pm 1]$ on this curve?
