Can an even perfect number be a sum of two cubes?

Question

A similar question was asked before in https://math.stackexchange.com/questions/2727090/even-perfect-number-that-is-also-a-sum-of-two-cubes, but no conclusions were drawn.

On the Wikipedia article of perfect numbers there are two related results concerning whether an even perfect number can be a sum of two cubes. Gallardo's result in 2010 (which can be found here) claims that 28 is the only perfect number that can be a sum of two cubes. This part is copied from the question on MSE, which is summarized from the paper:

Let $N$ be an even perfect number. Assume that $N=x^3+a^3=(x+a)(x^2-xa+a^2)$. Note that $x$ and $a$ have the same parity. Consider the case $x+a<x^2-xa+a^2$. By the Euclid–Euler theorem, it follows that $N=2^{p-1}(2^p-1)$, where $2^p-1$ is a Mersenne prime. Thus, $x+a=2^{p-1}$ and $x^2-xa+a^2=2^p-1$.

However, nowhere in the proof was it proven that both $x,a$ are odd, or that $x+a$ and $x^2-xa+a^2$ are coprime. If $x,a$ are even, the second equation cannot hold. So, is this result true? If the subsequent analysis is correct, this still shows that a perfect number cannot be expressed as two odd cubes. Or are there similar results concerning whether a perfect number can be expressed as a sum of two perfect powers?

Remark: the title of this paper is On a remark of Makowski about perfect numbers. The remark of Makowski, also referenced in the Wikipedia article, concerns the case $a=1$, so $x$ is also odd, and there is no issue of non-comprimality. For those interested, Makowski deduced that $x+1=2^{p-1}$ and $x^2-x+1 = 2^p-1$ from the fact that the latter factor must be odd. From these equations, $x=3$ follows immediately, hence $28$ is the only perfect number that is one more than a cube. @Mindlack's answer here generalizes the result to $N = x^m + 1$. Both proofs are elementary.

Answer 1

Here is a proof that 28 is the only even perfect number that is the sum of two positive cubes. The proof in Gallardo's article must be adapted in the case $x,a$ are even.

Write $N=2^{p-1}(2^p-1) = x^3+y^3 = (x+y)(x^2-xy+y^2)$. The gcd $d$ of $x$ and $y$ must be a power of 2, because $d^3$ divides $N$. Writing $x=2^h u$, $y=2^h v$ gives $2^{p-1-3h}(2^p-1) = u^3+v^3$. We are going to show that the only solution of the equation $2^k(2^p-1) = u^3+v^3$ with $k<p$ and $(u,v)=1$ has no solution for $p \geq 5$. First we must have $k \geq 1$ because $2^p-1$ is prime, and $u,v$ must be odd. Moreover $u+v=2^k$ and $u^2-uv+v^2=2^p-1$. Then \begin{equation*} 2^p-1=u^2-uv+v^2=u^2-u(2^k-u)+(2^k-u)^2 = 2^{2k}-3uv. \end{equation*} We deduce the bounds $p \leq 2k$ and \begin{equation*} 2^p-1 \geq 2^{2k}-3 \cdot (2^{k-1})^2 \geq 2^{2k-2}, \end{equation*} which implies $p \geq 2k-2$. Putting these bounds together, we get $p=2k-1$. So $u,v$ are solutions to the system of equations \begin{equation*} \begin{cases} & u+v = 2^k \\ & uv = (2^{2k-1}+1)/3 \end{cases} \end{equation*} The discriminant $\Delta$ of the polynomial $(X-u)(X-v)$ must be a perfect square. We have $\Delta=4 \cdot (2^{2k-2}-1)/3$, hence $2^{2k-2}-1$ is 3 times an odd square. In particular $2^{2k-2} - 1 \equiv 3 \bmod{8}$, which is possible only for $k=2$ and $p=3$.