<s>[INST] How to find an upper bound on the number of solutions of $y^3=x^2+4^k$ I have solved the first two parts of this question but I am struggling with the remaining section. I can't see any meaningful way to reuse what I did before and/or find a way forward.
Just to be clear it is part c) that I am stuck on.

Would anyone be able to help me thanks!
 [/INST] The following is heavily based on Theorem 3.3 in Keith Conrad's excellent "Examples of Mordell’s Equation".
$$y^2+4^k=x^3$$
Observe that $4^k$ is always a square, so we factor in $\mathbb{Z}[i]$
$$(y-2^ki)(y+2^ki)=y^2+2^{2k}=x^3$$
If $y-2^ki$ and $y+2^ki$ are cubes then: $y-2^ki=(m+ni)^3$, so $2^k=n(3m^3-n^2)$ there are only $2(k+1)$ possible factorization. Each factorization represent at most one possible solutions, so there are at most $2k+2$ solutions.
Look at the equation modulo 2: $y^2+4^k \equiv x^3 \pmod 2$ so $y \equiv x \pmod 2$.
Assume both $x$ and $y$ are odd and let $δ$ be a common divisor. So $δ$ also divides $(y+2^ki) - (y-2^ki)=2^{k+1}i$. Therefore, the norm of $δ$, $N(δ)$, divides $N(2^{k+1}i)=2^{k+1}$. However it also divides $N(y+2^ki)=y^2+4^k=x^3$ which is odd. Thus the norm of $δ$ is 1 and $(y+2^ki),(y-2^ki)$ are relatively prime. Because $\mathbb{Z}[i]$ is UFD, if a product of two relatively prime factors is a cube then the factors must also be cube up to units. Every unit in $\mathbb{Z}[i]$ is a cube so it can be absorbed inside the other factors and we don't need to worry about it. So $(y+2^ki),(y-2^ki)$ are indeed cubes.
Assume both x and y are even and let $x=2x',y=2y'$. The equation becomes $4{y'}^2+2^{2k}=8{x'}^3$, if k is greater than zero then we can divide by four and get ${y'}^2+2^{2k-2}=2{x'}^3$. If $k$ is greater than one, look modulo $2$ and get ${y'}^2\equiv 0 \pmod 2$ so $4|{y'}^2$ and thus $4|2{x'}^3$ and $2|x'$. Let $x'=2x'',y=2y''$, so $4{y''}^2+2^{2k-2}=16{x''}^3$. If $2^{2k-2}$ is not $1$, divide by $4$, get ${y''}^2+2^{2k-4}=4{x''}^3$. $y''$ must be divisible by $4$ if $2^{2k-4}$ is not one, let $y''=2y'''$ and divide by $4$. ${y'''}^2+2^{2k-6}={x''}^3$ continue like we started until the power of four get to one.
So we have $y^2=x^3-1,y^2=2x^3-1,y^2=4x^3-1$ as possible ending state. $y^2=2x^3-1$ is already dealt with in Conrad's notes (and this answer is getting long). In $y^2=4x^3-1$, $-1$ is not a quadratic residue modulo 4 and therefore has no solution. $y^2=x^3-1$ have only one integer solution and is reached when $k \equiv 0 \pmod 3$.
It is possible to do better - we can divide by $2^6$ as many time we want because we will get a rational point. $\frac{y^2}{2^6}+4^{k-3}=\frac{x^3}{2^6}$ becomes $\left(\frac{y}{2^3}\right)^2+4^{k-3}=\left(\frac{x}{2^2}\right)^3$. So we ask about rational points instead on that curve and the only thing that matters is $k \pmod 3$. 
$$
\begin{array}{clcr}
\text{Curve} & \text{Mordell-Weil group}& \text{Number of rational points} & \text{Easier bound} & \text{Overall bound} \\
\hline
y^2=x^3-2^{2\cdot0} & \mathbb{Z}/{2}\mathbb{Z} & 1 & 2k+3 & 1 \\
y^2=x^3-2^{2\cdot1} & \mathbb{Z} & \infty & 2k+2 & 2k+2 \\
y^2=x^3-2^{2\cdot2} & \text{Trivial} & 0 & 2k+2 & 0 \\
\end{array}
$$
See 0,1,2. I will admit I don't know how to prove these facts about Mordell-Weil groups at all yet.
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