Where is the flaw in this "proof" of the Collatz Conjecture?
There is a subtle issue with your induction argument: you are assuming that the Collatz conjecture holds for all integers $\leq n$, and then want to prove it holds for $n+1$ (strong induction). So far, so good.
You then prove that for some cases ($n+1$ even, or of the form $4k+1$) that the Collatz conjecture holds by the inductive hypothesis. Fine.
You then try to argue that for some numbers of the form $4k+3$, you eventually hit a number of the form $4k+1$, so that the Collatz conjecture holds... not so fast. You haven't proven that the Collatz conjecture holds for all integers of the form $4k+1$. You've proven it's true for $n+1$, if $n+1$ happens to be of that form, and you've assumed it's true for all numbers of that form $\leq n$ (by the inductive hypothesis) but you haven't shown that Collatz holds for numbers of the form $4k+1$ that are larger than $n+1$.
The error is:
We now know that a number $N+1$ can ONLY violate the Collatz Conjecture if $N$ is even and not a multiple of $4$.
We don't know that. What we know is that if there are numbers that violate the Collatz conjecture, then the smallest such number is $1$ more than something even and not a multiple of $4$. But there may be larger counterexamples which don't look like that.
This is how proof by induction works: if there's a counterexample, there must be a smallest counterexample, i.e. the conjecture is false for some number $n+1$ but true for all numbers $\leq n$. Then we can try to show that this can't happen. But if we can't deduce a contradiction from this, but only get some conditions which $n+1$ has to satisfy, then these conditions only hold for the smallest counterexample, not for every counterexample, since we had to assume that $n+1$ was the smallest counterexample in order to make the deduction.
In addition to the previous comments, there is another implicit assumption that needs to be addressed for this proof to work: The consideration for other loops.
The danger of this flaw is evident here: I used your method outlined in your proof, assuming $4k+1$ defines the numbers that converge to 1 and thus prove the Collatz Conjecture. I extended this formula to the modified Collatz rule $3x+5$, since the only difference is the $+5$ instead of $+1$. (I refrained from using $3x+3$ because its structure is completely different from that of $3x+1$ and $3x+5$.)
To achieve this, I used $4k+5$ for appropriateness, since the $n/2$ rule still applies for $3x+5$. I then look at the following sequence generated:
9,17,21,25,29,...
Immediately trouble unfolds. 9 goes to 32, which goes to 1, which completes the 8-4-2-1 loop.
13 goes to 11, which goes to 38, which then finds itself in the 19 loop. [19-62-31-98-49-152-76-38-19]
17 goes to 7, which goes to 13. 21 goes to 17 which goes to 7... maybe there is a pattern here?
25... ruins the entire new pattern. It goes to 80, which is a straight shot to the 5 loop [20-10-5-20].
And 29? Dead center of the 23 loop. [29-92-46-23-74-37-116-58-29].
Therefore, tuning this proof to prove that $3x+1$ will not wander to infinity is completely possible, however assuming every number reaches a power of 2 at some point may lead to further trouble. Especially since a sequence wandering to infinity for the Collatz Conjecture must never reach a power of 2 as $5x+1$ seems to demonstrate with its seemingly infinite 7 sequence.