Is $\sum_{n,m \geq 0} F_n^m x^n y^m$ a rational generating function?

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I am curious if the generating function defined by:

$$ F(x,y)=\sum_{n=0}^{\infty} \sum_{m=0}^{\infty} F_{n}^m x^n y^m$$

where $F_n$ is the $n$th fibonacci number, is a rational function. That is, Is $F(x,y) \in \mathbb{C}(x,y)$?

This is just out of curiosity, since the one variable case, where we fix either $m$ or $n$, always gives rise to a single variable generating function.

In my efforts to simplify, I have considered: $$ \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} F_{n}^m x^n y^m = \sum_{n=0}^{\infty} x^n \sum_{m=0}^{\infty} (F_{n}y)^m$$ $$ = \sum_{n=0}^{\infty}\frac{x^n}{1-F_ny}$$

But I cannot think of what to do from here.

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No, it grows too quickly. Set $y = x$ and you'll see that if $\sum a_{n, m} x^n y^m$ is rational then $a_{n, n}$ grows at most exponentially in $n$. But in your case $F_n^n$ grows like $\phi^{n^2}$.