An equation, where the solution does not exist, but on solving the equation we got a solution. why this is happening?

Let's consider a more simple example, to understand. Given the equation $$ x = 1 $$ you can take the square of both sides: $$ x^2 = 1 $$ and find two solutions: $$ x=1 \qquad x=-1. $$

This happens because the operation $x\mapsto x^2$ is not invertible. If you apply a non invertible function to an equation, the number of solutions might increase.

$-\frac{x}{x}= \frac{x}{x}$ has no solutions at all, since $-\frac{x}{x}\neq \frac{x}{x}$ no matter what $x$ is.

But we can square both sides, and then what happens?

$\frac{x^2}{x^2}= \frac{x^2}{x^2}$ is an equation that is true for all nonzero numbers.

By applying a non-invertible operation to both sides, we can turn an equation with no solutions into one with uncountably infinitely many solutions.

The process of solving an equation is basically that of inversion: you successively apply (inverse) functions to both sides of the equation until you reach a point where the solution is clear. This process depends on each successive equation (upon applying various inverses successively) being equivalent to the previous one so that the final equation $x=\ldots$ is equivalent to the original equation. However, when you apply non-invertible operations (such as $x\mapsto x^{2}$, i.e. squaring both sides), you don't get an equivalence between the equation before squaring and the equation after squaring: you get a forward implication, which is to say that the final equation $x=\ldots$ does not imply the previous equation(s) prior to squaring, it is only implied itself by the previous chain of equations.

Whenever we square, we immediately introduce extraneous root

Observe that $\displaystyle\frac54$ is actually a root of $$\sqrt{x+1}=\sqrt{4x-1}-\sqrt{x-1}$$

Also, observe that $\displaystyle2x-1=-2\sqrt{x^2-1}\le0\implies 2x\le1\iff x\le\frac12$ for real $x$

But, $\displaystyle{\sqrt{x-1}}$ is not real unless $x\ge1$

Write the equation as $$ \sqrt{x+1}=\sqrt{x-1}+\sqrt{4x-1} $$ Then you must have \begin{cases} x+1\ge0\\ x-1\ge0\\ 4x-1\ge0 \end{cases} which boils down to $x\ge1$. Now square, you're sure not to add spurious solutions, because both sides represent non negative numbers: $$ x+1=x-1+4x-1+2\sqrt{(x-1)(4x-1)} $$ or $$ -4x+3=2\sqrt{(x-1)(4x-1)} $$ Now the right hand side is non negative, so also the left hand side must be, which means $$ -4x+3\ge0 $$ or $x\le 3/4$. With the previous limitation, this has the consequence that no solution can exist.