= Solution
Antisymmetry gives $-F^{\rho\mu}\partial_\rho A^\nu=F^{\mu\rho}\partial_\rho A^\nu$. Combining this with the <canonical stress-energy tensor> removes the potential in favour of the <electromagnetic field tensor>:
$$
\begin{aligned}
\Theta^{\mu\nu}
&=-F^{\mu\rho}(\partial^\nu A_\rho-\partial_\rho A^\nu)
+\frac14\eta^{\mu\nu}F_{\rho\sigma}F^{\rho\sigma}\\
&=\boxed{-F^{\mu\rho}F^\nu{}_{\rho}+\frac14\eta^{\mu\nu}F_{\rho\sigma}F^{\rho\sigma}.}
\end{aligned}
$$
This is the <electromagnetic stress-energy tensor> in the $(+,-,-,-)$ convention. The product $F^{\mu\rho}F^\nu{}_{\rho}$ is symmetric in $\mu,\nu$, since the contracted metric is symmetric. Its expression only in $F$ makes it <gauge-invariant>. These two properties hold without using the equations of motion.
For conservation, the source-free <Maxwell equations> are $\partial_\rho F^{\rho\mu}=0$. Hence the given correction equals
$$
-F^{\rho\mu}\partial_\rho A^\nu=-\partial_\rho(F^{\rho\mu}A^\nu)
$$
on shell. This is an <antisymmetric superpotential improvement>: $B^{\rho\mu\nu}=-F^{\rho\mu}A^\nu$ is antisymmetric in $\rho,\mu$, so $\partial_\mu\partial_\rho B^{\rho\mu\nu}=0$. Conservation of the <canonical stress-energy tensor> therefore implies $\partial_\mu\Theta^{\mu\nu}=0$. Equivalently, direct differentiation gives $\partial_\mu\Theta^{\mu\nu}=-F^\nu{}_{\rho}\partial_\mu F^{\mu\rho}$ after the homogeneous <Maxwell equations> cancel the field-strength derivative terms.
There are four <conserved currents>, one for each fixed $\nu$, with charges $P^\nu=\int d^3x\,\Theta^{0\nu}$. The <Belinfante-Rosenfeld stress-energy tensor> differs from the canonical tensor only by boundary contributions to these charges when the fields decay suitably.
Finally, in four spacetime dimensions,
$$
\Theta^\mu{}_{\mu}=-F^{\mu\rho}F_{\mu\rho}+\frac44F_{\rho\sigma}F^{\rho\sigma}=0.
$$
Thus all requested properties are
$$
\boxed{\partial_\mu\Theta^{\mu\nu}=0\ \text{on shell},\qquad
\Theta^{\mu\nu}=\Theta^{\nu\mu},\qquad\delta_\xi\Theta^{\mu\nu}=0,\qquad\Theta^\mu{}_{\mu}=0.}
$$
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