I am trying to find rational solutions to the equation $x^y=y^x$, but I'm unsure if my solution is right (especially the first paragraph below). Here is my attempted solution (I've excluded the trivial case $x=y$).
We first show that any rational root must also be an integral root. To do so, consider $\mathcal F_b(x)=x^a-a^b$ for a parameter $b$ and constant $a$. By the rational root theorem, if $x=p/q$, $(p,q)=1$ is a rational root, then $q\mid 1$, hence $q=\pm 1$, and any root of $\mathcal F_b$ is integral, regardless of $b$. Since $x^y-y^x=\mathcal F_x(x)$, this conclusion applies as well.
Now we consider the case where $x,y>0$. Note that $x^y=y^x$ exactly when $y\ln x=x\ln y$ and $y/\ln y=x/\ln x$. Let $f(x)=x/\ln x$, so we have $f(x)=f(y)$. The derivative of this function is $f'(x)=(\ln x-1)/(\ln x)^2$, whose unique root is $x=e$. Thus on the open interval $(0,e)$, $f$ is monotonously decreasing, and on $(e,\infty)$ it is increasing. It is immediate that we cannot have $f(x)=f(y)$ if both $x$ and $y$ are on $(e,\infty)$ or $(0,e)$, thus exactly one of them is on $(0,e)$. Since they are positive integers, without loss of generality assume that $x>y$, so $y=1$ or $y=2$. If $y=1$ then $x=1$ but $x\neq y$, a contradiction. If $y=2$ then $x=4$. Likewise we also have the solutions $y=4$, $x=2$.
Then, we consider $x,y<0$. Then let us make the change of coordinates $x=-x'$ and $y=-y'$ with $x',y'>0$ to obtain $$ x^y=y^x\iff > (-x')^{-y'}=(-y')^{-x'} \iff (-x')^{y'}=(-y')^{x'}.$$ If exactly one of $x',y'$ is odd, then the signs on both sides are different, a contradiction. Hence they have the same parity. If both are even, then $(x',y')=(2,4),(4,2)$ since the equation reduces to the previous case; and hence $(x,y)=(-2,-4),(-4,-2)$. If both are odd, the same thing occurs, but neither of our two roots satisfy the supposition that both $x,y$ are odd. Hence there is no solution in this case.
Finally, we consider $x>0>y$ (the case $y>0>x$ is identically discussed). Considering signs at both sides of the equation $x^y=y^x$, we immediately have that $x$ is even. Make the change of coordinates $y=-y'$, and use that $x$ is even to get $x^{-y'}=(y')^x$. Notice that $(y')^x$ is a positive integer, hence $x^{-y'}$ is a positive integer as well. Thus $x^{-y'}\geq1\implies x^{y'}\leq 1$, and since $x,y'$ are positive integers, we must have $x=1$ or $y'=y=0$. If $x=1$, we have $1^y=y^1$, hence $y=1$. But $x\neq y$, a contradiction. If $y=0$, then $x^0=0^x$ and $1=0$, a contradiction. So there is no solution in this case.
Combining the three cases, our solutions are $(2,4),(-2,-4),(4,2),(-4,-2)$.
I'll restrict my comments to the first paragraph, which has a substantial gap. The rational root theorem applies to polynomials, requiring that powers of the variable be nonnegative integers. You can address this by inserting a step.
"Suppose $x = p/q$, $y = u/v$, with $p,q,u,v \in \mathbb{Z}$ and both $(p,q) = 1$ and $(u,v) = 1$, is a solution to $x^y = y^x$. Then $p,q,u,v$ are a solution to $((p/q)^{u/v})^{qv} = ((u/v)^{p/q})^{qv}$. More simply, $(p/q)^{qu} = (u/v)^{pv}$."
Now the exponents are rational, but we aren't quite done converting entirely to integers.
"Then we clear denominators: $q^{qu} v^{pv} (p/q)^{qu} = q^{qu} v^{pv} (u/v)^{pv}$, so $v^{pv} p^{qu} = q^{qu} u^{pv}$ has a solution in integers if $x^y = y^x$ has a solution in rationals."
You can attempt to continue as you have. The additional variables make it likely that you can find the rest of the many, many rational solutions. Note that at most one of $x$ or $y$ is negative, which allows you to place constraints on the signs of $p,q,u,v$ to ensure that $v^{pv} p^{qu} = q^{qu} u^{pv}$ has a non-negative power on whichever variable you choose to make the variable of your polynomial. (There will be a few cases, but not too many.) Skimming your argument, you will have to make several subsequent adjustments and I haven't dug into the details enough to see if your method can complete the argument.