= Solution
First assume global <four-dimensional N=1 supersymmetry>, canonical positive <Kähler potential>, and tree level. Write $W_i=\partial_i W$ and $m_{ij}=W_{ij}$ at the stationary vacuum. The <chiral-superfield fermion mass matrix> is the symmetric matrix $m$, and the sum of squared <Weyl spinor> masses is $\operatorname{tr}(m^\dagger m)$. From the <F-term scalar potential> $V=\sum_k|W_k|^2$, its mixed and holomorphic <Hessian matrix> blocks are
$$
V_{i\bar j}=\sum_kW_{ki}\overline{W_{kj}},
\qquad V_{ij}=\sum_k\overline{W_k}W_{kij}.
$$
For canonically normalized real and imaginary parts of the <scalar fields>, the real <scalar mass matrix> has trace twice the mixed trace. Its holomorphic blocks can split the two real scalar masses but do not change their sum. Hence
$$
\sum_{a=1}^{2n}m_{B,a}^2=2\operatorname{tr}(m^\dagger m),
\qquad 2\sum_{r=1}^{n}m_{F,r}^2=2\operatorname{tr}(m^\dagger m).
$$
A <Weyl spinor> has two spin states, so this proves the <tree-level supertrace mass sum rule>,
$$
\boxed{\operatorname{STr}M^2=\sum_am_{B,a}^2-2\sum_rm_{F,r}^2=0}.
$$
The paper defines its supertrace with $(-1)^{2j+1}$, the negative of the conventional boson-minus-fermion <supertrace> used here. The asserted zero is the same in either convention. The result requires the stated canonical tree-level hypotheses; a noncanonical <Kähler metric>, <supergravity>, or radiative corrections can change the sum rule.
If vector multiplets are also present, the <gauge contributions to the F-term supertrace> cancel rather than being omitted. At $D_a=0$, define $C=\sum_a g_a^2\|T_av\|^2$ for the scalar expectation vector $v$. Differentiating $\tfrac12\sum_aD_a^2$ adds $2C$ to the real-scalar trace. The symmetric <gaugino>-matter mass matrix has off-diagonal entries $\sqrt2g_aT_av$, giving an extra $4C$ in its fermion squared-mass trace. The covariant scalar <kinetic term> gives vector squared-mass trace $2C$. Thus their contribution is $2C-2(4C)+3(2C)=0$. The gauge-theory result uses all spin states, including the vector weight three; it is not obtained by dropping massive gauge partners.
The <MSSM tree-level sfermion mass constraint> explains the phenomenological difficulty with direct visible-sector breaking. If <F-term> breaking is neutral under an unbroken electric and color <gauge symmetry>, and no <D-term> shifts are present, the same trace argument applies within each conserved-charge fermion block. Its corresponding scalar partners cannot all have squared masses above the mean of the light fermion squared masses. Thus canonical tree-level visible-sector <supersymmetry breaking> alone cannot make every <squark> and other scalar partner heavy while leaving the observed fermions light. The usual effective <soft supersymmetry breaking> terms arise after communicating breaking from a separate sector; integrating out that sector, noncanonical interactions and radiative effects evade the hypotheses. The global trace identity alone would not identify a particular light <squark> without the conserved-charge block argument.
For the specified <O'Raifeartaigh model>, phases may be chosen so that $\lambda,\mu,M$ are real and positive. Its <superpotential> derivatives are
$$
W_X=\lambda(z^2-\mu^2),\qquad W_Y=Mz,\qquad W_Z=2\lambda xz+My,
$$
and its <F-term scalar potential> is
$$
V=\lambda^2|z^2-\mu^2|^2+M^2|z|^2+|2\lambda xz+My|^2.
$$
For fixed $x,z$, choose $y=-2\lambda xz/M$ to minimize the final square. Since $|z^2-\mu^2|^2\ge(|z|^2-\mu^2)^2$,
$$
V\ge\lambda^2\mu^4+(M^2-2\lambda^2\mu^2)|z|^2+\lambda^2|z|^4.
$$
Thus \b[$M^2>2\lambda^2\mu^2$] makes $z=y=0$ a global minimum with arbitrary $x$. The hierarchy $M\gg\mu$ ensures this for fixed perturbative $\lambda$, rather than for arbitrarily large $\lambda$. The origin is one member of this classically flat family, with $V_0=\lambda^2\mu^4>0$ and $F_X=\lambda\mu^2$. The complex field $x$ is a <pseudomodulus>.
At the origin the <chiral-superfield fermion mass matrix>, in the $(\psi_X,\psi_Y,\psi_Z)$ basis, is
$$
m=\begin{pmatrix}0&0&0\\0&0&M\\0&M&0\end{pmatrix}.
$$
Its physical fermion masses are \b[$0,M,M$]; the two massive <Weyl spinors> form one massive four-component fermion. The massless $\psi_X$ is the <goldstino>. More generally, stationarity gives $W_{ij}\overline{W_j}=0$, so the nonzero <auxiliary field> direction is a null vector of $m$, proving the <goldstino zero mode from vacuum stationarity>.
Writing $z=(z_R+iz_I)/\sqrt2$ and similarly for $x,y$, the quadratic <scalar potential> is
$$
V^{(2)}=M^2|y|^2+M^2|z|^2-\lambda^2\mu^2(z^2+\bar z^2).
$$
The <mass spectrum of the quadratic-cubic O'Raifeartaigh model> is therefore
$$
\boxed{\begin{array}{c|c}
\text{real scalar}&m^2\\\hline
x_R,x_I&0,0\\
y_R,y_I&M^2,M^2\\
z_R,z_I&M^2-2\lambda^2\mu^2,\ M^2+2\lambda^2\mu^2
\end{array}}.
$$
The scalar squared-mass sum is $4M^2$, while twice the fermion squared-mass sum is also $4M^2$, so \b[the supertrace is zero], as required. The two massless real $x$ modes are tree-level <pseudomoduli>, which can be lifted by a quantum effective potential; their tree-level masslessness is not the exact symmetry protection enjoyed by the <goldstino>.
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