Is there an easy proof of this equation related to simultaneous Pell equations?

Question

Working with the famous Baker-Davenport system of simultaneous Pell equations \begin{align} 3x^2-2 &= y^2, & 8x^2-7 &= z^2, \qquad(\star) \end{align} I am left, after a series of substitutions and reductions, with the equation $$ u(u+1)\left(96u(u+1)+11\right) = v(v+1). \qquad(\dagger) $$ The point is now to prove that the only solutions with $u \ge 0$ are $u=0$ and $u=5$, yielding, respectively, the two known (and only) solutions $x=1$ and $x=11$ in ($\star$).

Are there any ways of attacking the formulation ($\dagger$), different from those which would naturally be applied to ($\star$), which might lead to a relatively straight-forward proof?

Thanks,
Kieren.

EDIT: For example, by unique factorization of integers, we can write $(v,v+1)=(abc,rst)$ for postive integers $a,b,c,r,s,t$ with $gcd(abc,rst)=gcd(v,v+1)=1$. Now, with appropriate permutation, we can write \begin{align} u &= ar, & u+1 &= bs, & 96u(u+1)+11=ct. \end{align} But all methods and manipulations I've tried from that point seem to lead to dead ends.

Answer 1

Sorry, it's probably no easier: both are looking for integer points on a genus-$1$ curve. In fact it's not hard to get from Baker-Davenport to your equation: multiplying the two equations in $(\star)$ yields $(3x^2-2)(8x^2-7) = (yz)^2$, and both $x$ and $yz$ must be odd, so we can write $(x,yz) = (2u+1,2v+1)$, and then subtracting $1$ and dividing by $4$ yields $(\dagger)$.