Collatz Conjecture: Is there a straightforward argument showing that there are no nontrivial 2 "step" repeats (where each "step" is an odd number)

You got

$$x = \frac{3 + 2^{w_1}}{2^{w_1 + w_2} - 9} \tag{1}\label{eq1A}$$

For $x$ to be a positive integer means the denominator must divide evenly into the numerator which means the denominator must also be less than or equal to the numerator.

For somewhat easier algebra, let

$$y = 2^{w_1} \tag{2}\label{eq2A}$$

so \eqref{eq1A} now becomes

$$x = \frac{y + 3}{2^{w_2}y - 9} \tag{3}\label{eq3A}$$

This gives

$$y + 3 \ge 2^{w_2}y - 9 \iff 12 \ge (2^{w_2} - 1)y \iff \frac{12}{2^{w_2} - 1} \ge y \tag{4}\label{eq4A}$$

The denominator must also be a positive integer, but since the smallest power of $2$ greater than $9$ is $16$, we get

$$2^{w_2}y \ge 16 \iff y \ge \frac{16}{2^{w_2}} \tag{5}\label{eq5A}$$

Combining \eqref{eq4A} and \eqref{eq5A} gives

$$\frac{16}{2^{w_2}} \le y \le \frac{12}{2^{w_2} - 1} \tag{6}\label{eq6A}$$

For $w_2 = 1$, \eqref{eq6A} gives $8 \le y \le 12$. Since \eqref{eq2A} states $y$ is a power of $2$, the only possible solution is $w_1 = 3$ giving $y = 8$. However, \eqref{eq3A} gives $x = \frac{11}{7}$, which is not an integer.

Next, if $w_2 = 2$, \eqref{eq6A} gives $4 \le y \le 4$, i.e., $w_1 = 2$. Substituting these into \eqref{eq3A} gives $x = 1$.

If $w_2 = 3$, then \eqref{eq6A} gives $2 \le y \le \frac{12}{7}$, i.e., there's no value of $y$. Likewise, any value of $w_2 \gt 3$ will not allow any value of $y$. Also, Steven Stadnicki's question comment gives another way to see $w_2 = 2$ is its maximum possible value.

This means the only valid positive integer solution of \eqref{eq1A} is $w_1 = w_2 = 2$ giving $x = 1$.

I usually write it (if we have only few iterations I take letters $a,b,c,$) this way:

$$ b= {3a+1\over2^A} \qquad c= {3b+1\over2^B}\\ \text{ and } c=a \text{ to have a cycle} $$ One can rewrite this as $$ a \cdot b= {3a+1\over2^A} \cdot {3b+1\over2^B} \\ 2^S = \left( 3 + \frac1a\right) \cdot \left( 3 + \frac1b\right) $$ $ \qquad \qquad $ where $S=A+B$
Now we see, that the rhs is between $9$ (when $(a,b)=(\infty,\infty)$) and $16$ (when $(a,b)=(1,1)$) . So the the lhs, being a perfect power of $2$ must be $2^4=16$.
But the rhs to equal this $16$ we must have each parenthese equal to $4$ so we have proven:

To have a 2-(odd-)step cycle we need, that $(a,b)=(1,1)$ which represents also the trivial cycle.

Remark: one can extend this easily to 3-step-cycle et al. but for the 3-step-cycle we need also that $a \ne b \ne c$ and all $a,b,c \equiv \pm 1 \pmod 6$. Just try it!