Why is Coupled Cluster not variational?
Variational coupled cluster (vCC)
First, you can make a variational CC with a non-unitary T operator in the following way.
Recall the variational principle (taken from Wikipedia):

$$\varepsilon[\Psi] = \frac{\langle\Psi| \hat{H} |\Psi\rangle}{\langle\Psi|\Psi\rangle}$$
The variational principle states that

$\varepsilon \geq E_{0}$, where $E_{0}$ is the lowest energy eigenstate (ground state) of the hamiltonian
$\varepsilon = E_{0}$ if and only if $\Psi$ is exactly equal to the wave function of the ground state of the studied system.


So all you have to do to make a variational CC is set $|\Psi\rangle = e^{\hat{T}}|0\rangle$ in the above expression. This is called &quot;variational coupled cluster&quot; or &quot;vCC&quot;, because it literally uses the &quot;variational principle&quot; shown above.
Unitary coupled cluster (uCC)
Unitary transformations preserve eigenvalues, so if $\hat{U}$ is a unitary transformation (unlike $e^{\hat{T}}$), then $UHU^\dagger$ has the same eigenvalues as $H$. Since the energies are eigenvalues of $H$, unitary coupled cluster can recover the correct energies with the unitary operator $U = e^{\hat{T} - \hat{T}^{\dagger}}$. For a long time people thought vCC and uCC were just two ways of writing the same thing, but this was shown to be wrong very recently (G. Harsha et al. On the difference between variational and unitary coupled cluster theories. arXiv:1711.00579 [cond-mat.str-el]; J. Chem. Phys. 2018, 148 (4), 044107.).
Something that a lot of people don't know
Regular CC is actually variational in the full correlation limit. This means that for $\ce{Li2}$ for example, CCSDTQPH is variational. This is because $\ce{Li2}$ has $6$ electrons, so CCSDTQPH is FCI, and FCI is variational. CCSDTQP and CCSDTQ and even CCSDT(Q) are not likely to overshoot the variational limit either since they are very good approximations of CCSDTQPH for $\ce{Li2}$.