Let $f(i)$ and $g(j)$ be real and finite valued functions of positive integers $i$ and $j$, my numerical experiments suggest that If $$a_{ij}=|f(i)~g(j)|, A=[a_{ij}]_{n\times n}, \text{then}~~ Tr (A^k)=(Tr(A))^k,n,k \in I^+$$ I would like to see a proof for this observation, exceptions may also be pointed out.
I have tried all elementary transcendental and algebraic functions for $f$ and $g$.
You have $A = f g^T$ with vectors $f, g \in \Bbb R^n$. Therefore is $$ A^k = f (g^T f)^{k-1} g^T = (g^T f)^{k-1} \cdot f g^T = \left(\sum_{j=1}^n f_j g_j \right)^{k-1} \cdot f g^T $$ since $g^T f$ is a scalar. It follows that $$ \operatorname{Tr}(A^k) = \left(\sum_{j=1}^n f_j g_j \right)^{k-1} \operatorname{Tr}(f g^T) = \left(\sum_{j=1}^n f_j g_j \right)^k \, , $$ confirming your observation.
The fact that the entries are real-valued and non-negative is not needed, the same is true for real- or complex-valued vectors $f, g$, or for vectors in any commutative field $\Bbb K$.