Distinctions between Universal Generalization, Existential Instantiation, and Introduction Rule of Implication using an example claim

Question

From recent dives throughout these tags, I have learned that there are several different flavors of deductive reasoning (Hilbert, Genzten...natural deduction, sequent calculus...etc). This has made it a bit difficult to pick up on a single interpretation of how exactly Universal Generalization ("$\forall \text{I}$")$^1$, Existential Instantiation ("$\exists \text{E}$")$^2$, and Introduction Rule of Implication ("$\rightarrow \text{ I }$") $^3$ are different in their formal implementations.

Consider the following claim (which requires the the individual to carry out all of the three aforementioned inference rules):

$$\forall m \in \mathbb{Z} : \left( \exists k \in \mathbb{Z} : 2k+1 = m \right) \rightarrow \left( \exists k' \in \mathbb{Z} : 2k'+1 = m^2 \right)$$

In English: "For any odd number $m$, it's square is also odd".

For convenience let's have: $$\varphi(m):=\left( \exists k \in \mathbb{Z} : 2k+1 = m \right) \rightarrow \left( \exists k' \in \mathbb{Z} : 2k'+1 = m^2 \right)$$

By convention, the above statement is equivalent to the following:

$$\forall m \left[m \in \mathbb Z \rightarrow \varphi(m) \right]$$

In the following paragraphs, I will go through my understandings of this proof from purely the deductive argument side of things and sprinkle in the occasional explicit question, marked with a colored dagger ($\color{red}{\dagger}$). Any added commentary is greatly appreciated.

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As is typical with conditional based proofs, we say, "Assume $m^* \in \mathbb Z$". This is an application of ($\rightarrow \text{ I }$), and it establishes two things: 1) $m^*$ is now an unbound symbol representing something and 2) $m^*$ has the property that it is an integer.

Our goal is to then show that $\varphi(m^*)$ is true. Of note, $\varphi(m^*)$ is itself a conditional, and therefore we assume the antecedent of $\varphi(m^*)$, which is another invocation of ($\rightarrow \text{ I }$). We say, "Assume $\exists k \in \mathbb{Z} : 2k+1 = m^*$." This introduces another variable $k$, but I believe it is relevant to state that this new variable $k$ is bound, and therefore (I think) is not really a new variable in the sense that $m^*$ was ($\color{red}{\dagger}$).

As an aside, when I see existential claims, I think of sets whose elements satisfy the claim. For example, in the case of "$\exists k \in \mathbb{Z} : 2k+1 = m^*$", I think of the following set, which is non-empty by assumption: $S=\{k \in \mathbb Z \ |\ 2k+1=m^*\}$. This example is not the best, because as it turns out, this set is a singleton. However, one can easily envision a scenario where the set described by the existential claim is not-finite (i.e. countably or uncountably infinite)...in which case, it is not apparent to me at all why I am given license to "reach into this set" and pull an object out for the purpose of argument, as we will see next ($\color{red}{\dagger}$).

Now, by ($\exists E$), we say, "Choose a $k^* \in S$". We have just introduced a new symbol $k^*$ into our argument. Importantly, this symbol is unbounded. By definition of $S$, this means that $2k^*+1=m^*$.

Algebraic manipulation will subsequently reveal that:

\begin{align} (m^*)^2&=(2k^*+1)^2 \\ &=4(k^*)^2+4k^*+1 \\ &=2\left[(2k^*)^2+2k^* \right] +1 \\ \end{align}

It is easy to show that $(2k^*)^2+2k^*$ is itself an integer and satisfies the necessary property specified by the consequent. One then employs existential generalization to conclude $\exists k' \in \mathbb{Z} : 2k'+1 = (m^*)^2$.

Up to this point, we have shown that $m^* \in \mathbb Z \rightarrow \varphi(m^*)$. Taken from another post, here is the definition of ($\forall \text{ I }$)

If $P(c)$ must be true, and we have assumed nothing about $c$, then $\forall x P(x)$ is true.

In order to replicate the described form above, I suppose it is reasonable to collapse $m^* \in \mathbb Z \rightarrow \varphi(m^*)$ into a new formula $\psi(m^*):= m^* \in \mathbb Z \rightarrow \varphi(m^*)$. In which case, I would say that I proved $\psi(m^*)$. However, I most definitely did assume something about $m^*$. In fact, I assumed several things. Firstly, I assumed it is an integer. Secondly, I assumed that it satisfied that statement $\exists k \in \mathbb Z: 2k+1=m^*$. It seems to me that I have violated the conditions that would otherwise let me claim $\forall m \psi(m)$! ($\color{red}{\dagger}$)

The only thing I can think to do is create a new set $T = \{m \in \mathbb Z \ | \ \exists k \in \mathbb Z: 2k+1=m \}$. This set $T$ effectively represents the assumptions I have made. Then, I would argue I could claim: $\psi(m^*) \vdash \forall m \in T \left[\psi(m) \right]$.

Deconstructing what $\forall m \in T \left[\psi(m) \right]$ means, we effectively have the form:

$\forall m \left [ A \land B \rightarrow \left(A \rightarrow \left(B \rightarrow C \right) \right) \right]$, which I am relieved to find out is equivalent to simply $\forall m \left [A \rightarrow (B \rightarrow C) \right]$...i.e. $\forall m \psi(m)$.

I have never seen the above work carried out in any post/article/book, perhaps because, in the end, it does not matter. If I could have confirmation that this is correct thinking, I would greatly appreciate it ($\color{red}{\dagger}$).

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