Consequence of the Noether normalization lemma [duplicate]

Here is the more algebraic solution you're asking for. First, we should require that our base field is infinite - we could bypass this requirement if we were working in the affine/non-graded case (see for instance a very nice proof by Mel Hochster here which works over any field and in fact generalizes to arbitrary domains), but the structure of the argument we're using here does require this.

Let $k$ be an infinite field and let $R$ be a graded $k$-algebra which is finitely generated by $x_0,\cdots,x_n$ in degree one. Then there are $y_0,\cdots,y_m\in R$ so that each $y_i$ is a $k$-linear combination of the $x_j$, the $y_i$ are independent, and $R$ is finite over $k[y_0,\cdots,y_m]$. We prove this by induction on $n$: the base case where $R$ is generated by $x_0$ is clear, as either $R$ is a polynomial ring in $x_0$ or $R$ is finite over $k$.

For the inductive step, let $\varphi:k[x_0,\cdots,x_n]\to R$ be the obvious graded surjection. If $\ker\varphi=0$, we're done. Else, let $P\in \ker\varphi$ be a nonzero homogeneous element of degree $d$. I claim that up to a change of variables of the form $x_i=x_i+\lambda_ix_n$ and $x_n=\lambda_nx_n$, we may assume $P=x_n^d+\cdots$: writing $P$ as a polynomial in $x_0$ and expanding out after this change of variables, we have $$P=P(\lambda_0,\cdots,\lambda_n)x_n^d+\cdots$$ where the $\cdots$ represents terms of degree less than $d$ in $x_n$. As any nonzero polynomial over an infinite field has a nonzero value, we can find $\lambda_i$ so that $P(\lambda_0,\cdots,\lambda_n)\neq 0$ and therefore $k[x_0,\cdots,x_n]/(P)$ is finite over $k[x_0,\cdots,x_{n-1}]$ and surjects on to $R$. Now consider the image of $\varphi(k[x_0,\cdots,x_{n-1}])$ in $R$: by the previous sentence, $R$ is finite over this, and by induction, we can find an appropriate selection of $y_i$ so that $\varphi(k[x_0,\cdots,x_{n-1}])$ is module-finite over $k[y_0,\cdots,y_m]$. As the composition of finite extensions is finite, we're done.