The Fermat-Catalan conjecture with signature $(2,n,4)$, $n\ge4$ The Fermat-Catalan conjecture is that for coprime $x,y,z$ and positive integers $a,b,c$ with $1/a+1/b+1/c<1$, the generalized Fermat equation $x^a + y^b = z^c$ has only finitely many solutions. I'm considering signatures $(a,b,c)$ which are solved.
Table 1 of [BCDY] surveys known results and states that $(2,n,4)$, $n\ge4$ has been solved completely and that this is 'Immediate from Bennett–Skinner [BS], Bruin [Br3]'. [Br3] covers the case $n=5$. Fermat dealt with $n=4$.
This leaves $n=6, 9$ and prime $n\ge7$, but I can't see how [BS] is relevant to that. Can someone explain and/or point me to the relevant part of [BS].
[BCDY] 'Generalized Fermat equations: A miscellany', Bennett, Chen, Dahmen, Yazdani, International Journal of Number Theory, Vol. 11, No. 1 (2015)
[BS] 'Ternary Diophantine Equations via Galois Representations and Modular Forms', Bennett, Skinner, Canad. J. Math. Vol. 56(1), 2004 p23-54.
[Br3] 'Chabauty methods using elliptic curves', Bruin, J.reine angew. Math. 562 (2003), 27-49.
Note: This question was originally posted in MSE on 2020-07-03. It's had some upvotes, but no answers as of 2020-08-24.
 A: [Br2] Theorem 1 covers the case $n=6$. So this leaves $n=9$ and prime $n\ge7$.
As suggested in a comment, I contacted Michael Bennett directly and he kindly explained the rest to me:
We have $x^2+y^n=z^4$ with $x,y,z$ coprime integers.
So $(z^2 - x)(z^2 + x) = y^n$.
The gcd of $(z^2 - x)$ and $(z^2 + x)$ is $1$ or $2$.
For a gcd of $1$ we have $x$ and $z$ of opposite parity, and can write
$z^2-x = u^n$ and $z^2+x = v^n$
so that $u^n+v^n = 2z^2$.
This is solved for coprime integer $u,v,z$ for $n\ge4$ by [BS] Theorem 1.1.
For a gcd of $2$ we have $x$ and $z$ both odd, and one of
$z^2-x = 2 u^n$ and $z^2+x = 2^{n-1}v^n$, or
$z^2+x = 2 u^n$ and $z^2-x = 2^{n-1}v^n$.
In either case, $u^n + 2^{n-2} v^n = z^2$.
This is solved for coprime integer $u,v,z$ for prime $n\ge7$ by [BS] Theorem 1.2.
This leaves the case $n=9$ with $y$ even.
Going back to the original equation, we have
$x^2+y^9=z^4$.
[Co, Section 14.4.1] gives complete parametrizations of
$x^2+w^3=z^4$
in terms of $s$ and $t$. In our case, $w$ is an even cube and from this and the parity constraints on $s$ and $t$ given in [Co] it follows that there exist coprime integers $s$ and $t$ with
$s t (s^3 - 16 t^3) (s^3 + 2 t^3)$ a cube.
The factors on the left hand side are pairwise coprime. ($s^3-16t^3$ and $s^3+2t^3$ could possibly have a common factor of $3$, but if they do, the whole expression is divisible by $9$ but not by $27$, and hence is not a cube.)
Since the factors are pairwise coprime, $s^3+2t^3$ is a cube.
This corresponds to a rational point on the curve $A^3+2B^3$=1 which is isomorphic to the elliptic curve $Y^2=X^3-1728$ via standard transformations. The latter curve has rank $0$ (and only the rational points corresponding to the point at infinity and $(X,Y)=(12,0)$). Tracing these back to $A^3+2B^3=1$, we find that $(A,B)=(1,0)$ or $(A,B)=(-1,1)$.
These points lead to either $t=0$ (which gives $z=0$ in $x^2+y^9=z^4$) or to $st = -1$ (which does not make $s t (s^3 - 16 t^3) (s^3 + 2 t^3)$ equal to a cube).
[Br2] 'The Diophantine Equations $x^2 \pm y^4 = \pm z^6$ and $x^2 + y^8 = z^3$', Bruin, Compositio Mathematica 118: 305-321, 1999.
[Co]  'Number Theory Volume II: Analytic and Modern Tools', Henri Cohen
