Radius of convergence of $\sum_{n=0}^{\infty}2^nx^{n^2}$
I will use the Root test, let $c_n = 2^nx^{n^2}$
$\lim_{n\rightarrow \infty} c_{n}^{1/n} = \lim_{n\rightarrow \infty} 2x^{n} = 0 $ if $-1<x<1$ and $\pm \infty$ if $|x|>1$
Since we want $\lim_{n\rightarrow \infty} c_{n}^{1/n}<1$ for convergence as according to the Root test, so the radius of convergence must be $1$. Is this correct?
Your reasoning is correct: however you can calculate the radius of convergence of the series directly, by using the definition, i.e. for a given power series $$ \sum_{m=1}^\infty c_m x^m \quad\text{ we have }\quad \limsup_{m\to \infty}\sqrt[m]{|c_m|}=\frac{1}{R} $$ But if you chose this way, you should be careful: precisely, you should consider the coefficients of the powers of $x$ which are different from $0$ and calculate the their root respect to the exponent of their associated power. Considering your series this means $$ c_m= \begin{cases} 2^n&\text{ if }m=n^2\\ 0 & \text{ otherwise} \end{cases} $$ Thus $$ \begin{split} \frac{1}{R}&=\limsup_{m\to \infty}\sqrt[m]{|c_m|}=\limsup_{n\to\infty} \sqrt[n^2]{2^n}\\ &=\limsup_{n\to\infty}2^\frac{n}{n^2}=\limsup_{n\to\infty}2^\frac{1}{n}\\ &=\limsup_{n\to\infty} \sqrt[n]{2} =1 \end{split} \implies R=1 $$