Here's a somewhat different approach. First, similar to what you did, the "if" part means each prime factor of $m$ is congruent to $1 \pmod{4}$. As shown in the answer to Sum of two squares and prime factorizations, Fermat's theorem on the sum of squares states each prime factor $p_i$ of $m$ can be written as the sum of squares. Also, for any $c, d, e, f \in \mathbb{R}$,
$$(c^2 + d^2)(e^2 + f^2) = (ce \pm df)^2 + (cf \mp de)^2 \tag{1}\label{eq1A}$$
shows whenever $2$ numbers can be written as a sum of squares, their product can be as well, in $2$ different ways. Using \eqref{eq1A} repeatedly with the previous result (starting at $1$) and for each $p_i \mid m$ means the final product, i.e., $m$, can be written as a sum of squares.
Regarding proving you can choose an $a$ and $b$ where $\gcd(a, b)$, the answer to Any product of primes in the form of 4n+1 is the sum of 2 relatively prime squares shows this, paraphrased below.
As shown in \eqref{eq1A}, the product of the $2$ sums of squares can be expressed in $2$ ways. Have $c^2 + d^2$, with $\gcd(c, d) = 1$, be a product of $1$ or more primes of the form $4n + 1$, and $e^2 + f^2$ be a prime of that form to be multiplied. Consider if the first form in \eqref{eq1A}, i.e., $(ce + df)^2 + (cf - de)^2$, is not valid, i.e., there's a prime $q$ which divides each term. This means
$$q \mid (ce + df)e + (cf - de)f = c(e^2 + f^2) \tag{2}\label{eq2A}$$
$$q \mid (ce + df)f - (cf - de)e = d(e^2 + f^2) \tag{3}\label{eq3A}$$
Since $q$ doesn't divide $c$ and $d$, then $q \mid e^2 + f^2 \implies q = e^2 + f^2$. If both solution types in \eqref{eq1A} are not valid, then $e^2 + f^2$ divides $ce - df$ as well as $ce + df$, and hence divides $2ce$ and $2df$. Since $e^2 + f^2$ doesn't divide $2e$ or $2f$, it must divide both $c$ and $d$, contrary to the hypothesis, meaning at least one of the $2$ forms must be valid. Thus, use the valid form, and repeat this procedure for each prime that is multiplied, to eventually get $m$.
For the "only if" part, similar to the answer to If $a \in \Bbb Z$ is the sum of two squares then $a$ can't be written in which of the following forms?, suppose there's a prime $p \equiv 3 \pmod{4}$ with $p \mid m$. If $p \mid a$, then $p \mid b$, and vice versa, but since $\gcd(a, b) = 1$, then $p$ can't divide either $a$ or $b$. Thus, $a$ has a multiplicative inverse, call it $a'$, modulo $p$. Let $r = \frac{p-1}{2}$ and note $r$ is odd. Also using Fermat's little theorem, this gives (note the argument below is basically equivalent to showing $-1$ is not a quadratic residue modulo $p$ if $p \equiv 3 \pmod{4}$)
$$\begin{equation}\begin{aligned}
a^2 + b^2 & \equiv 0 \pmod{p} \\
a^2(a')^2 + b^2(a')^2 & \equiv 0 \pmod{p} \\
1 + (ba')^2 & \equiv 0 \pmod{p} \\
(ba')^2 & \equiv -1 \pmod{p} \\
\left((ba')^2\right)^{r} & \equiv (-1)^r \pmod{p} \\
(ba')^{p-1} & \equiv -1 \pmod{p} \\
1 & \equiv -1 \pmod{p}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
This, of course, is not possible, meaning the original assumption must be false. This confirms all prime factors of $m$ must be congruent to $1 \pmod{4}$.
Turns out the reference for part reduction was a book by Schinzel, he of the Hypothesis.
there is a nice answer to this sort of thing. Throw in a variable $z$ to homogenize,
$$ f = x^3 + A x z^2 + B z^3 + C y z^2 + y^3 $$
Next, find the second partial derivatives of $f$ by $x,y,z$ and write out the Hessian matrix. The entries in $H$ are linear in $x,y,z,$ as they begin degree three.
Finally, calculate the determinant of $H.$ This is a homogeneous degree three. The form $f$ factors into linear terms over $\mathbb C$ if and only if the determinant is a constant multiple of $f.$ The reference is by Brookfield, I believe I saved part of it, let me look.
Gary Brookfield (2016) Factoring Forms, The American Mathematical
Monthly, 123:4, 347-362
Best Answer
Your idea is good. It can be formalized in a clearer way.
For a finite subset $F$ of $I$, define $d(F)$ to be the gcd of the members of $F$. It is easy to show that if $F_1\subseteq F_2$, then $d(F_2)$ is a divisor of $d(F_1)$.
Then there exists $G$ such that $d(G)$ is minimal. I contend that $d(G)$ is $1$. Indeed, if $p$ is a prime divisor of $d(G)>1$, we can find $b\in I$ such that $p\nmid b$, otherwise every element of $I$ would be divisible by $p$.
Then $p\nmid d(G\cup\{b\})$, so $d(G\cup\{b\})$ is a proper divisor of $d(G)$, hence smaller. Contradiction.
Finally, the submonoid generated by $G$ is contained in the submonoid generated by $I$. Since the former contains all positive integers from some point on, the same is true for the latter.