<s>[INST] What is the most elementary way of proving a sequence is free of non-trivial squares? Given the sequence A001921
$$
0, 7, 104, 1455, 20272, 282359, 3932760, 54776287, 762935264, 10626317415, 148005508552, 2061450802319, 28712305723920, 399910829332567, \dots
$$
which obeys the recurrence relation
$$
  a_0 := 0,\ a_1 := 7, \quad a_{n+2} = 14a_{n+1} - a_n + 6.
$$
How can I prove (or, I suppose, disprove) my conjecture that $a_0 = 0$ is the only square?
In case it helps, the conjecture is equivalent to saying that $x$ and $3x^2+3x+1$ cannot both be [positive integer] squares, though I am particularly interested in any general method of proof that attacks the recurrence relation directly.
EDIT: Here's an application.
 [/INST] Theorem. If $a$ and $b$ are integers such that $3a^4+3a^2+1=b^2\!$, then $a=0$.
Proof. Assume, contrary to the claim, that $a \ge 1$ and $b>1$ are integers satisfying the equation. Evidently $b$ is odd, say $b=2c+1$ for an integer $c \ge 1$. Hence
\begin{align*}
 3a^4 + 3a^2+1 &= (2c+1)^2  \\
 3a^2(a^2+1) &= 4c(c+1).
\end{align*}
Since $4 \nmid 3(a^2+1)$, regardless of the parity of $a$, we must have $2 \mid a^2$; hence $a$ is even, say $a=2d$ for an integer $d \ge 1$. Now
\begin{align*}
 3d^2(4d^2+1) &= c(c+1).
\end{align*}
We now show that this equation has no solutions with $c,d \ge 1$. Assume to the contrary that there exist integers $p,q,r,s \ge 1$ such that
\begin{align*}
 3d^2 &= pq, &&&
  c &= pr,  \\
 4d^2+1 &= rs, &&&
  c+1 &= qs.
\end{align*}
Adding and factoring gives
\begin{align*}
 3d^2+c+1 &= pq+qs = q (p+s),  \\
 4d^2+c+1 &= rs+pr = r(p+s),
\end{align*}
and then by subtraction we obtain $(p+s)(r-q) = d^2\!$. Since $3d^2=pq$, we have either $p \mid 3$ or $\gcd(p,d)>1$. The latter together with $(p+s) \mid d^2$ would contradict $\gcd(p,s) \mid \gcd(pr,qs) = \gcd(c,c+1) = 1$. Hence $p \mid 3$, so $p=1$ or $p=3$.
Case 1: $p=1$. Then $q=3d^2$, and $c=r$. Now $c+1 = qs$, so by substitution $r+1 = 3d^2s$. On the other hand, $rs-1 = 4d^2$, and adding these two relations yields
\begin{align*}
 (rs-1)+(r+1) &= 4d^2 + 3d^2s  \\
 r(s+1) &= d^2(3s+4).
\end{align*}
Since $3s+4=3(s+1)+1$ implies $\gcd(s+1,3s+4)=1$, and $rs=4d^2+1$ implies $\gcd(r,d)=1$, we conclude $r=3s+4$ and $s+1=d^2$. Hence
\begin{align*}
 4d^2+1 &= rs  \\
  &= \bigl(3(d^2-1)+4\bigr)(d^2-1)  \\
 2 &= 3d^2(d^2-2),
\end{align*}
which is clearly impossible.
Case 2: $p=3$. Then $q=d^2$ and $c=3r$. Now $c+1=qs$ becomes $3r+1=d^2s$. Since it is still true that $rs-1=4d^2$, we add the two relations to obtain
\begin{align*}
 (rs-1) + (3r+1) &= 4d^2 + d^2s  \\
 r(s+3) &= d^2(4+s).
\end{align*}
Again, by considering common factors, we conclude $r=s+4$ and $s+3=d^2$. Hence
\begin{align*}
 4d^2+1 &= rs  \\
 &= (d^2+1)(d^2-3)  \\
 d^2(d^2-6) &= 4.
\end{align*}
This is also impossible, completing the proof.
</s>