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Can $$ X_t - a X_{t-1} = Z_t + Z_{t-1} $$ be written with recursion as $$ X_t - a^2 X_{t-2} = Z_t + (1-a) Z_{t-1} $$ where $a = \begin{pmatrix} 0 & 1/3 \\ 1/3 & 0 \end{pmatrix} $?
I wasn't sure of how to find the autocovariance when the MA(1) did not have a companion matrix.
No, it generally cannot, unless $Z_{t-2} = 0$ for all $t$: Define the lag operator $L$ such that $LX_t = X_{t-1}$, and similarly for $Z_t$. Then the equation you start with above can be rewritten as a lag polynomial of form \begin{align} (1-aL)X_t = (1-L)Z_t. \end{align} Using the third binomial formula, it is obvious that $(1+aL)(1-aL) = (1-a^2L^2)$, and as $(1-a^2L^2)X_t = X_t - a^2X_{t-2}$ by definition of the lag operator, one can now write \begin{align} (1+aL)(1-aL)X_t & = (1+aL)(1-L)Z_t\\ (1-a^2L^2)X_t & = (1 + (a-1)L - aL^2)Z_t \\ X_t - a^2X_{t-2} & = Z_t + (a-1)Z_{t-1} - aZ_{t-2}. \end{align} So in particular, $X_t - a^2X_{t-2} = Z_t + (1-a)Z_{t-1}$ if and only if $(a-1)Z_{t-1} - aZ_{t-2} = (1-a)Z_{t-1}$, which obviously only holds if $Z_{t-2} = 0$.
