Prove that there are no integers $x$ and $y$ such that $3x^2=13+4y^2$

$3x^2=\underbrace{13+4y^2}_\text{odd}$

We know that x is odd lets' pose $x=2k+1$

\begin{align}3(2k+1)^2=13+4y^2\\ 3(4k^2+4k+1)=13+4y^2\\ 3(4k^2+4k)=10+4y^2\\ 12(k^2+k)=10+4y^2\\ \underbrace{6(k^2+k)}_\text{even}=\underbrace{5+2y^2}_\text{odd} \end{align}

but $5+2y^2$ is odd not even which contradict the factor 6 on the left side of the equation

You can look mod4.

Your equation is equivalent to

$1=3x^{2} \pmod 4$.

$3=x^{2}\pmod 4$.

But 3 is not a square mod 4.

Square of an integer is always of the form $4k$ or $4k+1$.

$$\text{RHS} =4(y^2+3)+1=4n+1$$

While, on the other hand (left one! :P)

$$\text{LHS}=3(4m) ~\text{or} ~3(4m+1) \equiv 4l ~ \text{or} ~4l+3$$

LHS and RHS both leave different remainders while dividing with $4$, therefore they can never be equal.