* Let $T(n, k)$ be [A107670][1] (i.e., matrix square of triangle [A107667][2]). Here we define the triangular matrix $P$ by $P(n, k) = \frac{(n+1)^{2(n-k)}}{(n-k)!}$ for $0 \leqslant k \leqslant n$ and the diagonal matrix $D$ by $D(n, n) = n+1$ for $n \geqslant 0$; then $T$ is given by $T = P^{-1}D^2P$.
* Start with vector $\nu$ of fixed length $m$ with elements $\nu_i = 1$ (that is, $\nu = \{1, 1, \dotsc, 1\}$), then set $t$ as an empty vector of fixed length $m-1$ and for $i$ from $1$ to $m-1$ and for $j$ from $2$ to $i+1$ apply $\nu_j := \nu_j + iz\nu_{j-1}$ and $t_i = i^2\nu_i$ (after ending each cycle for $j$).

I conjecture that after the whole transform we have vector $t$ with elements $t_i=\sum\limits_{k=0}^{i-1}T(i-1,k)z^{i-k-1}$.

Here is the PARI/GP program to check it numerically:

    upto1(n) = my(v1); v1 = vector(n+1, i, 1); v2 = vector(n, i, 0); for(i=1, n, for(j=2, i+1, v1[j] += i*z*v1[j-1]); v2[i] = i^2*Vec(v1[i])); v2
    row1(n) = my(P=matrix(n+1, n+1, r, c, if(r>=c, (r^2)^(r-c)/(r-c)!)), D=matrix(n+1, n+1, r, c, if(r==c, r))); vector(n+1, i, (P^-1*D^2*P)[n+1, i])
    test1(n) = upto1(n) == vector(n, i, row1(i-1))

Is there a way to prove it?

  [1]: https://oeis.org/A107670
  [2]: https://oeis.org/A107667