Edited to add: Well, now I feel embarrassed to have gotten an answer accepted which is absolute garbage, so I think I should offer an actual answer in addition to the indirect proof in a comment I made above (i.e. the Betti numbers of the cokernel $A$ of your matrix are 2,2,4,11,32,95,..., while those of the residue field $k$ are 1,4,13,40,121,364,..., so $k$ can't be a direct summand of $A$).
Here's another that doesn't rely on computer algebra software. It does rely on $A$ being a graded module over the (naturally) graded ring $R$. Suppose $A \cong N \oplus k$. It's easy to check that $A$ has hilbert function $(2,6,1)$. Since $A$ is generated in a single degree, the copy of $k$ must also be generated in that degree, so $N$ has hilbert function $(1,6,1)$. In particular $N$ must be cyclic, $N \cong R/J$ for some $J$. But $R$ has hilbert function $(1,4,3)$, so can't have a quotient with hilbert function $(1,6,1)$.
Edit: ThisThe below is wrong. Sorry.
The minimal generators of the module $\mathrm {coker}\ M$ are the column vectors $(v,x)^T$ and $(y,z)^T$. They generate a two-dimensional vector space of all the minimal generators of the module. This is just $X/mX$, where $X = \mathrm{coker}\ M$ and $m=(x,y,z,v)$. If there is going to be a direct summand isomorphic to $k$, there must be a minimal generator which is annihilated by the maximal ideal. But one can write down a generic minimal generator $(av+by, ax+bz)^T$ and the 8 $k$-linear equations saying that it is annihilated by $x,y,z$ and $v$. Two of them are $v(av+by)=0$ and $z(ax+bz)=0$. The relations in the ring imply $bvy=0=avx$. Since $vy$ and $vx$ are nonzero in $R$, this means $a=0=b$, and so there is no such direct summand.
Would you tell us why you thought there should be such a direct summand?