The problem is
Prove that $$3x^{10} - y^{10} = 1991$$ has no integral solutions.
I have written this proof, of which I am not sure.
Assume that $x$ and $y$ are integers. $$\begin{align} 3 x^{10}-y^{10} &\equiv 2 \mod 3 \\ \implies \qquad y^{10} &\equiv 2 \mod 3 \\ \implies \qquad y^{10} &= 3 m + 2, \text{where } m\in\mathbb{Z} \\[8pt] \therefore \qquad 3x^{10}-y^{10} &=3x^{10}-3m-2=1991 \\[8pt] \implies 3 \left(x^{10}-m\right) &= 1993 \end{align}$$
But $3\not\mid 1993$, $1993 = 3k+1$, where $k\in\mathbb{Z}$. $$\begin{align} \implies& 3\left(x^{10}-m\right) = 3k+1 \\ \implies& x^{10}-m \;\text{is not an integer} \end{align}$$
This contradicts our assumption that $x^{10}$, $m$ are integers, because if they were then $x^{10}-m$ would also be an integer, which [it] is not. Hence, our assumption is wrong, therefore the above equation does not have any integral solutions.
Please read my proof. If there are any errors, then please help in resolving them. Any other method of proof is also appreciated. But please tell if my proof is correct.
Your proof fails at the very first step; you wrote down the implication $$3 x^{10}-y^{10} \equiv 2 \pmod{3}\qquad \implies \qquad y^{10} \equiv 2 \pmod{3},$$ which is false. This leads to your contradiction later in the proof.
Replacing it by the true implication $$3 x^{10}-y^{10} \equiv 2 \pmod{3}\qquad \implies \qquad y^{10} \equiv 1 \pmod{3},$$ does not lead to a contradiction later in the proof. So this approach will not work.
As suggested in the comments, working modulo $11$ is much more natural because it allows you to use Fermat's little theorem; you have $x^{11}\equiv x\pmod{11}$ for all $x$.