A numerical matrix of power sum polynomials

Question

Let $p_i=x_1^i+x_2^i+\cdots+x_m^i=\sum_{k=1}^mx_k^i$ be the power sum polynomials. Then, the determinant of the $m\times m$ Hankel matrix $M_m=(p_{i+j-2})$, for $1\leq i,j\leq m$, has a neat evaluation $$\det M_{\color{blue}m}=\prod_{1\leq a<b\leq {\color{blue}m}}(x_b-x_a)^2.$$

Notice the number of variables $x_k$'s and the dimension of the matrix are equal. As soon as this changes, the determinant lacks it elegance in the present general form. However, if we are willing to scale down the task to numerical entries (from $x_k$ to $k$) then experiments show an encouraging signal.

QUESTION. Is this true? $$\det\left(\sum_{k=1}^{{\color{red}{n+m}}}k^{2i+2j-2}\right)_{i,j=1}^{{\color{blue}m}} =m!^2\prod_{1\leq a<b\leq m}(b^2-a^2)^2\prod_{1\leq i\leq j\leq 2m}\frac{2n+i+j}{i+j}.$$

Postscript 1. Thanks to Ira Gessel's response, I went on to check further and observed that the two determinants, that of Zavrotsky $\det(S_p^{i+j})$ and ours $\det(S_p^{2i+2j})$, maintain nice evaluations. On the other hand, higher powers such as $\det(S_p^{3i+3j})$ do not enjoy the same crisp values. We used the notation $S_p^i=1^i+2^i+\cdots+p^i$.

Postscript 2. Following Sam Hopkins' great effort on references for the product term $\prod_{1\leq i\leq j\leq 2m}\frac{2n+i+j}{i+j}$, we may add the following:

S. R. Ghorpade, C. Krattenthaler, The Hilbert series of Pfaffian rings. – Appendix: Geometry of degeneracy loci and a plethora of multiplicity formula, ArXiv 2001 (see page 12, Theorem 2).

Jakob Jonsson Generalized triangulations and diagonal-free subsets of stack polyominoes, Journal of Combinatorial Theory, Series A 112 (2005) (see page 130, Corollary 17).

Z. Hamaker, N. Williams, Subwords and Plane Partitions, Daejeon, South Korea DMTCS proc. FPSAC’15 (see page 244, Theorem 6).

R. A. Proctor New Symmetric Plane Partition Identities from Invariant Theory Work of De Concini and Procesi, Europ. J. Combinatorics V. 11 (1990) (see page 292, Theorem 1, CGI).

R. P. Stanley Symmetries of plane partitions, J. Comb. Theory, Series A, V. 43, Issue 1, (1986) (see page 107, Case 6).

Answer 1

A similar determinant, with $k^{i+j-2}$ instead of $k^{2i+2j-2}$ is evaluated in A. Zavrotsky, El Gesseliano (Spanish), Notas de Matematicas, no. 73, Universidad de los Andes, Facultad de Ciencias, Departamento de Matematica, Merida, Venezuela, 1985. The evaluation can also be found in my paper with Harald Helfgott, Enumeration of Tilings of Diamonds and Hexagons with Defects, Electronic J. Combin. 6 (1999), Article R16, Lemma 11.

Here it is for the convenience of the reader. Denote $S_p^i=\sum_{k=1}^pk^i$ and $V_n=\prod_{k=1}^nk!$. Then, $$\det(S_{n+m}^{i+j-2})_{i,j=1}^m=\frac{V_{n+2m-1}V_{n-1}V_{m-1}^4}{V_{n+m-1}^2V_{2m-1}}.$$

Perhaps the same approach will work for this determinant.