Infinite Multiplication $=$ Infinite Addition

Which sequence? The LHS or the RHS? I can't answer your question properly otherwise.

For the RHS, Herschfeld's convergence theorem.

Another way to think of the LHS is to think of it as an addition of powers.

$\sqrt{5\sqrt{5\sqrt{5\sqrt{5\sqrt{\cdots}}}}} = 5^\frac{1}{2} \cdot 5^\frac{1}{4} \cdot 5^\frac{1}{8} \cdot 5^\frac{1}{16} \cdots$

$\large 5^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots}$

$\Large 5^{\sum_{n = 0}^{\infty}{\frac{1}{2} \cdot \left(\frac{1}{2}\right)^n}}$

$\sum_{n = 0}^{\infty}{\frac{1}{2} \cdot \left(\frac{1}{2}\right)^n} = 1$

$5^1 = 5$

In other words, no, the LHS can never be equal to zero unless the number inside is zero itself. As to why I put it as a possible answer in my solution; I put it in as you could calculate it without needing to know how to use infinite sums, thus making it easier to calculate in this case.

If you were trying to make $z$ equal to the RHS of the equation then you've made a small mistake.

$\sqrt{x^2 + 2\sqrt{x^2 + \sqrt{x^2 + \sqrt{x^2 + \cdots}}}}$

$\sqrt{x^2 + \sqrt{4x^2 + 4\sqrt{x^2 + \sqrt{x^2 + \cdots}}}}$

$\sqrt{x^2 + \sqrt{4x^2 + \sqrt{16x^2 + 16\sqrt{x^2 + \cdots}}}}$

$\sqrt{x^2 + \sqrt{4x^2 + \sqrt{16x^2 + \sqrt{256x^2 + \cdots}}}}$

As you can see when expanding your version you end up excluding $64x^2$ from the nested radical, in other words:

$\small \sqrt{x^2 + 2\sqrt{x^2 + \sqrt{x^2 + \sqrt{x^2 + \cdots}}}} \neq \sqrt{x^2 + \sqrt{4x^2 + \sqrt{16x^2 + \sqrt{64x^2 + \sqrt{\cdots}}}}}$

My advice would be to check more terms before assuming you have the answer. Other than that however, the rest of your solution was algebraically sound.

I want to know how you came up with this method, Animesh Singh ?

I started with with the addition of two variables; $a$ and $b$ and expanded from there.

$a + b = \sqrt{(a + b)^2}$

$a + b = \sqrt{a^2 + 2ab + b^2}$

$a + b = \sqrt{a^2 + \sqrt{(2ab + b^2)^2}}$

$a + b = \sqrt{a^2 + \sqrt{4a^2b^2 + 4ab^3 + b^4}}$

$a + b = \sqrt{a^2 + \sqrt{4a^2b^2 + \sqrt{(4ab^3 + b^4)^2}}}$

$a + b = \sqrt{a^2 + \sqrt{4a^2b^2 + \sqrt{16a^2b^6 + 8ab^7 + b^8}}}$

$a + b = \sqrt{a^2 + \sqrt{4a^2b^2 + \sqrt{16a^2b^6 + \sqrt{(8ab^7 + b^8)^2}}}}$

$a + b = \sqrt{a^2 + \sqrt{4a^2b^2 + \sqrt{16a^2b^6 + \sqrt{(64a^2b^{14} + 16ab^{15} + b^{16}}}}}$

$a + b = \sqrt{a^2 + \sqrt{4a^2b^2 + \sqrt{16a^2b^6 + \sqrt{64a^2b^{14} + \sqrt{(16ab^{15} + b^{16})^2}}}}}$

$a + b = \sqrt{a^2 + \sqrt{4a^2b^2 + \sqrt{16a^2b^6 + \sqrt{64a^2b^{14} + \sqrt{\cdots}}}}}$

I then made $b$ equal to $1$ and $a$ equal to $x$

$x + 1 = \sqrt{x^2 + \sqrt{4x^2 + \sqrt{16x^2 + \sqrt{64x^2 + \sqrt{\cdots}}}}}$

As far as I know there's no other way to go about it.