= Solution
In vacuum, curl <Faraday's law> and use $\nabla\cdot E=0$ and the <Ampère-Maxwell equation> to obtain
$$
\nabla^2E-\frac1{c^2}\partial_t^2E=0,\qquad c=(\mu_0\varepsilon_0)^{-1/2}.
$$
The plane wave solves this when $\omega=c|k|$ and $k\cdot E_0=0$. <Faraday's law> gives
$$
B=\operatorname{Re}\left(\frac{k\times E_0}{\omega}e^{i(k\cdot x-\omega t)}\right).
$$
For the stated wave, propagation is along $+x$ and polarization is $(0,1,1)/\sqrt2$. Thus
$$
B=\frac{E_0}{c}(0,-1,1)\frac{\cos(kx-\omega t)}{\sqrt2},
$$
$$
S=\frac{E_0^2}{\mu_0c}\cos^2(kx-\omega t)e_x,\qquad
\langle S\rangle=\frac{E_0^2}{2\mu_0c}e_x.
$$
The <Poynting vector> is electromagnetic energy flux, and its average is the wave intensity.
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