How to show that the ring $R= \mathbb Z_3[x]/\langle x^6-1\rangle$ is finite without using the concepts of vector space? I have understood that since $\mathbb Z_3$ is a finite field so $R$ has to be a finite dimensional vector space but is there any way of proving this from purely ring theoretic concepts ?

Hint $\ $ Let $\,R = S[x]/f(x)$ where $\,f\,$ is not constant $\,f\not\in S,\,$ and $\,f\,$ is monic (lead coeff $= 1).\,$ By the Polynomial Division Algorithm, every $\,g\in S[x]\,$ is congruent mod $f$ to its remainder $r$, which has degree lower than $\,\deg f,\,$ i.e.
$$ g  = q f + r,\ \ \deg r< \deg f\ \Rightarrow\ g\equiv r \pmod f$$
If, further, the coefficient ring $\,S$ is finite then there are only finitely many polynomials of such bounded degree, so only finitely many elements in $\,S[x]/f(x).$ 
For example, if $\,f\,$ is a monic quadratic, then every element of $S[x]/f$ is congruent to a linear polynomial $\,ax+b,\,$ a fact familiar from $\,\Bbb C \cong \Bbb R[\,i\,]\cong \Bbb R[x]/(x^2\!+1),\,$ where every element has the form $\,a\,x+b\pmod{\!x^2\!+1},\,$ i.e. $\,a\,i+b\,$ where $\,i := x\pmod{\!x^2\!+1},\,$ i.e. $\ x + (x^2\!+1)\Bbb R[x].$ 
Remark $\, $ In your case the Polynomial Division Algorithm is slightly overkill, since we can use the Integer Division Algorithm on exponents $\, \color{#c00}{x^6\equiv 1}\ \Rightarrow\ x^n\! = x^{6j+k}\!\equiv (\color{#c00}{x^6})^j x^k\equiv x^k\ $ for $\,k\le 6$.
