I think I have a plausible explanation. You're probably familiar with the generator for Pythagorean triplets:

$x^2-y^2=a \quad 2xy=b \quad x^2+y^2=c \quad \longleftrightarrow \quad a^2+b^2=c^2$

One could say this is a complete solution to the equation $a^2+b^2=c^2$ in the sense that 1) $x$ and $y$ can be chosen randomly to calculate $a,b,c$, and 2) for every example of $a^2+b^2=c^2$ where $a$ is coprime to $b$, there will always be $x$, $y$ such that one of $x^2-y^2$ or $2xy$ is $a$ and the other is $b$.

Along a similar line, complete solutions to $a^2+2b^2=m^i$, for positive $i$, can be obtained by generating two sequences of polynomials $f_1,f_2,f_3$... and $g_1,g_2,g_3$... by

$f_1 := x$

$g_1 := y$

$f_{i+1} := x(f_i)-2y(g_i)$

$g_{i+1} := x(g_i)+y(f_i)$

In other words, if you find positive coprime $a$ and $b$, and $a^2+2b^2=m^i$, there will always be $x$ and $y$ such that $x^2+2y^2=m$, $\left\vert f_i \right\vert = a$, and $\left\vert g_i \right\vert = b$.

I can prove this, and would suppose it is already well known. I generally find after any discovery that some mathematician already treated the subject more comprehensively and elegantly a few centuries ago. In any case, my proof starts with a few things proven directly, but the final part of it is by infinite descent. [3/31/2017 - Since my earlier post, I have noticed that a substantial part of what leads to the final infinite descent argument can also be accomplished with infinite descent. I would say I can rigorously prove that 27 is the only cube two greater than a square using "ninety percent" infinite descent arguments, and "ten percent" direct algebraic and logic arguments.] By that time, I have established that if there is an example of $a^2+2b^2=m^i$, with no corresponding $x$ and $y$, then $m$ must be a composite number. Furthermore, if $m$ is divisible by the prime number $p$, there will be another counter-example for $m_2 := {m \over p}$. Then there is an infinite sequence $m, m_2, m_3$, ... and every time $m_i$ is divided by a prime number, the result is $m_{i+1}$ which is composite.

I can provide further details if need be. It seems to me at least possible that Fermat used this approach. Certainly it leads with ease to the solution to $c^2+2=m^3$. In this case $a=c, b=1, i=3$. Since 1 is coprime to all numbers, $a$ is coprime to $b$. Then $f_3$ and $g_3$ should generate the example.

$$f_3 = x^3-6xy^2 = \pm c$$
$$g_3 = 3x^2y-2y^3 = \pm 1$$
$$x^2+2y^2=m$$

In the middle equation, 1 is a multiple of $y$, and the only possibilities for
$y$ are 1 or -1. That leads to the only solutions being $x=\pm1, y=\pm1$, and all four of these lead to the same end result, $c=5, m=3$.

The equation $c^2+4=m^3$ can be treated in a similar manner, by extending the Pythagorean triplet generator. In this case use $f_{i+1} := x(f_i)-y(g_i)$ instead of what is given above, but keep the definition for the $g$'s. However, there is a slight additional complication. Solving $g_3=\pm 2$ will lead to the example $11^2+4=5^3$ as well as the example $2^2+4=2^3$ in which $a$ is not coprime to $b$. However, 2 is not coprime to all numbers, and you will need to additionally demonstrate that 2 is the only even value for $c$ such that $c^2+4$ is a cube. Note the stipulation that $a$ is coprime to $b$ is necessary. For example, $9^2+12^2=15^2$ has no $x,y$ because 15 is not the sum of two squares.