Since $A$ is a real symmetric matrix, it is diagonalizable over $\mathbb{R}$.
So there exist real eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_r$
such that
$\operatorname{tr}(A^n)=\lambda_1^n+\lambda_2^n+\cdots+\lambda_r^n\qquad \text{for all } n\in\mathbb{N}.$
Let
$\rho=\max{1\le i\le r} |\lambda i|.$
We consider three cases.
$\textbf{Case 1: } \rho<1$
Then $\lambda_i^n\to 0$ for every $i$, hence $\operatorname{tr}(A^n)\to 0.$
But the condition requires $|\operatorname{tr}(A^n)|\ge 2^{\sqrt n},$ and $2^{\sqrt n}\to\infty$.
This is impossible.
$\textbf{Case 2:} \rho=1$.
Then $|\lambda_i^n|\le 1$ for every $i$, so $|\operatorname{tr}(A^n)| \le |\lambda1^n|+\cdots+|\lambda r^n| \le r.$
Thus $(|\operatorname{tr}(A^n)|)$ is bounded.
But again $2^{\sqrt n}\to\infty,$
so this is impossible.
$\textbf{Case 3: } \rho>1$.
Let the eigenvalues with absolute value $\rho$ be $\pm \rho.$
For even powers $n=2m$, we have $\lambda i^{2m}=|\lambda i|^{2m}.$
Hence every eigenvalue with $|\lambda_i|=\rho$ contributes exactly $\rho^{2m}$.
So if $k\ge 1$ is the number of eigenvalues satisfying $|\lambda_i|=\rho$, then
$\operatorname{tr}(A^{2m}) = k \rho^{2m} + \sum{|\lambda_i|}$
Therefore, $\operatorname{tr}(A^{2m}) \sim k\rho^{2m}\qquad (m\to\infty).$
So $|\operatorname{tr}(A^{2m})|$ grows exponentially like $\rho^{2m}$.
But the required upper bound is
$|\operatorname{tr}(A^{2m})| \le 2020\cdot 2^{\sqrt{2m}},$ and $2^{\sqrt{2m}}$ grows subexponentially.
An exponential function cannot be bounded above by a subexponential one for all large $m$.
This is impossible.
Since all three cases lead to contradictions, no such symmetric matrix $A$ can exist.
$\textbf{Conclusion:}$ The statement is $\boxed{\text{False}}.$
