Let $E$ be a field extension of a field $F$.
Noether's Normalization Lemma states (in this case) that if $E = F[y_1,...,y_n]$, that is E is a finitely generated $F$ - algebra, then there exists an algebraically independent (over F) set $\{x_1,...,x_k\}$ in $E$, such that $E$ is integral over $F[x_1,...,x_k]$.
The weak Nullstellensatz tells us that in this situation, $E/F$ is a finite extension.
Here is the the source of my confusion: If $\{x_1,...,x_k\}$ is algebraically independent, then is it not contradictory that the dimension of the field extension is finite? For example, take $x_1$, and assume the dimension is finite. Since the extension is finite, if we take powers of $x_1$, eventually we get a linearly dependent set of vectors over F. But then we have some polynomial in $x_1$ that is zero. But this contradicts the algebraic independence of the set of $x_i$.
I'm sure I have made a simple incorrect assumption somewhere, I would appreciate any help finding it!
In general, the Noether Normalization Lemma states that for a commutative $F$-algebra $A$ over a field $F$, there exists a $k \geq 0$ and algebraically independent elements $x_1, \dots, x_k \in A$ such that $A$ is a finitely generated as a $F[x_1,\dots,x_k]$-module.
When $A$ is a domain, $k$ is the transcendence degree of (the field of fractions of) $A$ over $F$.
So, indeed, in your case $A = E$ is an algebraic field extension of $F$, so it has transcendence degree $0$ over $F$.