<s>[INST] Does this system of simultaneous Pell-like equations have any non-trivial positive integer solutions? Let $a,b,c$ be positive integers satisfying
\begin{align}
2a^2-1 &= b^2,  \\
2a^2+1 &= 3c^2.
\end{align}
The trivial solution is $(a,b,c)=(1,1,1)$. Are there others?
 [/INST] 
NOTE: As pointed out by Erick Wong, there was a flaw in the original proof. It may be fixable, though I haven't found a fix yet; leaving here for history (and just in case anyone else can take this method to the goal line).

Theorem. The Diophantine system of equations
\begin{align}
2a^2-1 &= b^2, \\
2a^2+1 &= 3c^2
\end{align}
has only the trivial solution $\lvert a\rvert = \lvert b \rvert = \lvert c \rvert = 1$.
Proof. Evidently, $\lvert a\rvert = \lvert b \rvert = \lvert c \rvert = 1$ is a solution. We also see that $\lvert c \rvert = 1$ forces $\lvert a\rvert = \lvert b \rvert = \lvert c \rvert = 1$. We now assume, contrary to the theorem, that there is a solution with $bc > 1$ [positivity assumed without loss of generality].
Multiplying the two original equations yields
\begin{align}
  (2a^2-1)(2a^2+1) &= (b^2)(3c^2)  \\
  (2a^2)^2 - 1 &= 3(bc)^2.
\end{align}
Evidently $bc$ is odd, say $bc=2u+1$ for an integer $u \ge 1$ [because $bc > 1$]. There are exactly two cases to consider.
Case 1: $a \equiv 1\!\pmod{3}$. Say $a=3v+1$ for an integer $v \ge 1$ [because $a > 1$]. Substituting, expanding, and simplifying yields
\begin{align}
  \bigl(2(3v+1)^2\bigr)^2 - 1 &= 3(2u+1)^2  \\
  324v^4 + 432v^3 + 216v^2 + 48v + 3 &= 12u^2+12u+3  \\
  v(3v+2)(9v^2+6v+2) &= u(u+1).
\end{align}
Since all the factors are positive [by hypothesis], the Fundamental Theorem of Arithmetic implies there exist positive integers $p,q,r,s$ such that
\begin{align}  \tag{1}
v(3v+2) &= pq, &
u &= pr, \\
9v^2+6v+2 &= rs, &
u+1 &= qs.
\end{align}
The left-hand pair of relations implies $rs-3pq=2$, and the right-hand pair implies $qs-pr=1$. Solving these two equations for $q$ and $r$ in terms of $s$ and $p$ yields
\begin{align} \tag{2}
q &= \frac{s+2p}{s^2-3p^2}, &
r &= \frac{2s+3p}{s^2-3p^2}.
\end{align}
Since $q$ and $r$ are both integers [by hypothesis], it must be that $s^2-3p^2$ divides both $s+2p$ and $2s+3p = 2(s+2p)-p$, and so it must divide $p$. But $qs-pr=1$ implies $\gcd(p,s)=1$, and hence $\gcd(s^2-3p^2,p)=1$. Therefore we conclude $s^2-3p^2 = \pm 1$; since $s^2-3p^2=-1$ has no integer solutions, we have $s^2-3p^2=1$. Substituting this into (2) gives $q=s+2p$ and $r=2s+3p$.
[PROOF INCORRECT AFTER THIS POINT]
NOTE: One can quickly deduce that $(s,p)$ and $(r,q)$ are consecutive solutions to $U^2-3V^2=1$, with $r>s$ and $q>p$. Since $(2a^2,bc)$ is also a solution, we can let $(s,p)$ be the $n$th solution, so that $(r,q)$ is the $(n+1)$th, and $(2a^2,bc)$ is the $(2n+1)$th. One can then show [using standard elementary Pell theory] that $(b,c)=(q+p,s+p)$, so that $b-c=2p$. From that point, I can't get any more useful information with this method.
</s>