By contradiction, if $\pi^e$ were rational, then we could write $\pi^e=\frac{a}{b}$ where $a,b\in\mathbb{I}^+$ and $b\neq0$. So: $$\begin{align} \\ \pi^e&=\frac{a}{b} \\ e\ln(\pi)&=\ln(a)-\ln(b) \\ e\int_1^\pi \frac{1}{t}dt&=\int_1^a \frac{1}{t}dt-\int_1^b \frac{1}{t}dt \\ e&=\frac{\int_b^a \frac{1}{t}dt}{\int_1^\pi \frac{1}{t}dt} \end{align}$$ But we know that $e$ is irrational. So if $p=\int_b^a \frac{1}{t}dt$ and $q=\int_1^\pi \frac{1}{t}dt$, there can't exist an $a$ and $b$ such that $\frac{p}{q}=e$. Therefore by contradiction, $\pi^e$ must be irrational. The proof is similar for $e^\pi$. Q.E.D.
Does this work, or did I assume something I couldn't?
This proof is not correct. The fact that $e$ is irrational means that you can't write $e=\frac{p}{q}$ where $p$ and $q$ are both integers. Your $p$ and $q$ are not integers (at least not obviously so), so you don't get a contradiction. Every number $x$ can be written as a fraction $\frac{p}{q}$ for some $p$ and $q$ (for instance, $x=\frac{x}{1}$); this does not mean every number is rational.