$\DeclareMathOperator{\tr}{tr}$For a real, square matrix $A$, I believe that one has a simple upper bound on the (absolute value of the) trace of its square in terms of the trace of its Gramian-type product, namely \begin{align} \left|\tr{A^2}\right| &= \left|\sum_i \lambda_i(A^2)\right|\\ &\leq \sum_i \left|\lambda_i(A^2)\right|\\ &\leq \sum_i \sigma_i (A^2) = \tr\left(A^T A\right), \end{align} where $\lambda_i(\,\cdot\,)$ and $\sigma_i(\,\cdot\,)$ denote the $i$th largest eigenvalue and singular value, respectively. [The equality at the end was obtained by noting that the singular values of $A^2$ are equal to the eigenvalues of $((A^2)^TA^2)^{1/2} = A^T A$.]
My question is, are there any methods or theorems to establish a lower bound on $\tr{A^2}$ in terms of $\tr\left(A^T A\right)$ (potentially with constants, factors of its dimension/rank, or square roots)? (In particular, I am dealing with a class of matrices that satisfy $\tr{A^2} = \left(\tr{A}\right)^2$, and where $\tr{A}$ is strictly real, so I am equally interested in bounds on simply $\tr{A}$, if an inequality exists for that as well.)
$\DeclareMathOperator{\tr}{tr}$You can't get a nontrivial lower bound on $|\tr(A^2)|$, if that's what you meant, because it is quite possible to have $\text{tr}(A^2) = 0$ while $\text{tr}(A^T A) \ne 0$. For example, try $$A = \pmatrix{1 & -1\cr 1 & 1\cr}$$
As for $\tr(A^2)$ itself, from your bound you have
$$ \tr(A^2) \ge - \tr(A^T A) $$
with equality e.g. for
$$ \pmatrix{0 & -1\cr 1 & 0\cr}$$