A finite dimensional algebra over a field has only finitely many prime ideals and all of them are maximal
A finite dimensional (commutative or not) algebra over a field is an artinian ring, because it certainly satisfies the descending chain condition on left or right ideals.
Let $A$ be a commutative artinian ring and let $P$ be a prime ideal. Then $A/P$ is an artinian domain, which is a field: if $D$ is an artinian domain, consider $d\in D$ and the chain of ideal $(d)\supseteq(d^2)\supseteq\dots\supseteq(d^n)\supseteq\dotsb$. This chain stabilizes, so $(d^n)=(d^{n+1})$ for some $n$, which means that $d^n=td^{n+1}$. Since $D$ is a domain, $1=td$ or $d=0$. Therefore all prime ideals of $A$ are maximal.
Consider now a minimal element in the family of products of finitely many maximal ideals, let it be $M_1\cdots M_k$. If $M$ is a maximal ideal, then $$ MM_1\cdots M_k\subseteq M_1\cdots M_k $$ so, by minimality, $MM_1\cdots M_k=M_1\cdots M_k$ and $$ M_1\cdots M_k\subseteq M $$ Since $M$ is prime, it follows that $M_i\subseteq M$, for some $i=1,\dots,k$; therefore $M_i=M$.
Here we don't use any general fact about artinian rings, just that any nonempty set of ideals has a minimal element (which is general theory of lattices). The Chinese remainder theorem can be used to get an upper bound on the number of maximal ideals; let $M_1,\dots,M_k$ be the set of (pairwise distinct) maximal ideals; then the Jacobson radical $J(A)=M_1\cap\cdots\cap M_k=M_1\cdots M_k$ by coprimality; so $$ A/J(A)\cong A/M_1\times\cdots\times A/M_k $$ by the CRT. If $A$ is a finite dimensional $K$-algebra, then $k\le\dim_K A/J(A)\le\dim_K A$. The upper bound can be reached: the product algebra $K^n$ has exactly $n$ maximal ideals.
Now that the mild ambiguity is resolved, let's address the question.
We begin by showing maximality. Consider a non-trivial prime ideal, $\mathfrak{p}$ of $R$. If $x\not\in\mathfrak{p}$ is fixed, then considering the $\mod \mathfrak{p}$ reductions of $1, x,\ldots, x^{\dim_K(R)}=x^m$ we have a linear dependence by finite dimensionality of $R$. And since $R/\mathfrak{p}$ is an integral domain, we know that this algebraic relation
$$c_0+c_1x+\ldots +c_mx^m$$
has $c_0\ne 0$, otherwise $\overline{x}\in R/\mathfrak{p}$ would be a zero divisor, so that the reduction, $\overline{1}\in R/\mathfrak{p}$ satisfies
$$\overline{1}=c_0^{-1}(-c_m\overline{x}^m-\ldots-c_1\overline{x})$$
verifying the existence of an inverse for $x\mod \mathfrak{p}$, since
$$\overline{1}=\overline{x}\cdot\bigg(-c_0^{-1}(c_m\overline{x}^{m-1}+c_{m-1}\overline{x}^{m-2}+\ldots + c_2\overline{x}+c_1)\bigg)$$
If there are no more than $\dim_K(R)$ prime ideals we are done, so assume otherwise. Then we take a collection, $\{\mathfrak{p}_1,\ldots,\mathfrak{p}_n\}$, of $n=\dim_K(R)+1$ distinct prime ideals of $R$. Since all prime ideals are maximal, $\mathfrak{p}_i+\mathfrak{p}_j=R$ when $i\ne j$. Then by the CRT we can find $x_1,\ldots, x_n$ such that
$$x_k\equiv\delta_{ik} \mod \mathfrak{p}_i,\quad 1\le k\le n$$
Necessarily the $x_i$ span the quotient $K$-algebra
$$R/(\mathfrak{p}_1\cdot\ldots\cdot\mathfrak{p}_n)\cong R/\mathfrak{p}_1\oplus\ldots\oplus R/\mathfrak{p}_n$$
treated as a vector space, which is of dimension at least $\dim_K(R)+1$, implying there is a surjective $K$-algebra homomorphism from $K^m\to K^{M}$ for some $m<M$. However algebra homomorphisms are also linear maps, which means we have a vector space of lower dimension surjecting onto one of higher dimension. Hence there are at most $\dim_K(R)$ prime ideals, i.e. finitely many.
Remarks ($1$)In fact any such system of congruences over the finitely many prime ideals has $\gcd(\{x_i\})=1$, i.e. the $x_i$ are coprime.
($2$) Really we could have stopped after solving the system of congruences, as it already implies that the $x_i$ are linearly independent over $K$. To see this we can assume, WLOG, that
$$x_1=\sum_{i=1}^{\dim_K(R)+1}r_ix_i$$
Then $x_1\in \left(x_2,\ldots, x_{\dim_K(R)+1}\right)$, since the latter is an algebra ideal, a contradiction to the congruences.