It seems to me you are confusing two methods to compute the radius of convergence of the power series $\sum_{n=1}^\infty (\sqrt{n} - 1)^{\sqrt{n}} z^n$.
(1) The root test is used to determine when a series (not necessarily a power series, which has a variable) converges. See https://en.wikipedia.org/wiki/Root_test
In this case, if we choose some $\tilde{z} \in \mathbb{C}$, then the power series converges in $z = \tilde{z}$ is the series $\sum_{n = 1}^\infty (\sqrt{n} - 1)^{\sqrt{n}} \tilde{z}^n$ converges. The series has terms $c_n = (\sqrt{n} - 1)^{\sqrt{n}} \tilde{z}^n$ and if the limit $C = \lim_{n \to \infty} \sqrt[n]{|c_n|}$ exists, then the series converges if $C < 1$ and diverges if $C > 1$. In our case, the limit does exists and similar to how you did, you may rewrite it as $$C = \lim_{n \to \infty} (\sqrt{n}-1)^{1/\sqrt{n}} |\tilde{z}| = |\tilde{z}| \lim_{n \to \infty} (\sqrt{n}-1)^{1/\sqrt{n}} = |\tilde{z}| \cdot 1 = |\tilde{z}|.$$ We see that the power series always converges in $z = \tilde{z}$ with modulus $<1$ and does not converge in $z = \tilde{z}$ with modulus $>1$, so the largest disc in which the power series converges, has radius $1$. Hence, the radius of convergence is 1.
(2) You can also use a criterion to directly calculate the radius of convergence, see
https://en.wikipedia.org/wiki/Radius_of_convergence#Finding_the_radius_of_convergence
This method uses the root test and says that, if we write the power series as $$\sum_{n = 1}^\infty a_n z^n, \quad a_n = (\sqrt{n} - 1)^{\sqrt{n}},$$ then the radius of convergence is $R = 1/L$, with $$L = \limsup_{n \to \infty} \sqrt[n]{|a_n|}.$$ If the limit $\lim_{n \to \infty} \sqrt[n]{a_n}$ then it is equal to the limsup (this is so in our case) and then $$L = \lim_{n \to \infty} \sqrt[n]{a_n} = \lim_{n\to \infty}(\sqrt{n} - 1)^{1/\sqrt{n}}.$$ Note that there are no $z$'s or $\tilde{z}$'s in this expression! We have calculated this limit to be $1$ and so $R = 1/1 = 1$ indeed.
A quick note about the limit $L = \lim_{n \to \infty}(\sqrt{n} - 1)^{1/\sqrt{n}}.$ Because $\sqrt{\cdot}$ is increasing, we can indeed find that this limit equals $L = \lim_{m \to \infty} (m - 1)^{1/m}.$ For large $m$, we have $$(m-1)^{1/m} \approx m^{1/m}$$ so we intuit $L = \lim_{m\to \infty} m^{1/m} = 1$. (See e.g. How to show that $\lim_{n \to +\infty} n^{\frac{1}{n}} = 1$?)
We prove the limit with the sandwich theorem. Remark that for $m \geq 2$ we have $1 \leq m - 1$ and thus $1 \leq (m-1)^{1/m}$. On the other hand, we also have $(m-1)^{1/m} \leq m^{1/m}$. So, for all $m \geq 2$ we have the following sandwich $$1 \leq (m-1)^{1/m} \leq m^{1/m}$$ and since the outer limits $1$ and $m^{1/m}$ both have limit $1$ as $m \to\infty$, we can happily conclude $L = 1$.
