How do we know when no square matrix can have RRE form and CRE form that is given? Defining column echelon form as the transpose of reduced row echelon form, how could one go about find a square matrix given its row reduced echelon form (RREF) and column reduced echelon form (CREF)? How do we know when no square matrix can have RRE form and R form that is given?
For example can a matrix have RREF $M_1$, CREF $M_2$,
$$M_1=
\begin{pmatrix}
1&0&-1\\
0&1&3\\
0&0&0
\end{pmatrix}
\;M_2=
\begin{pmatrix}
1&0&0\\
2&0&0\\
0&1&0
\end{pmatrix}$$
The rowrank and columnrank respectively are equal but what method can deduce the form or lack of existence?
 A: 
Claim: Given a matrix $M_1$ in RREF and $M_2$ in CREF where both matrices have rank $r$, the matrix
$$
M = M_2M_1
$$
will be such that $M_1$ is the RREF of $M$ and $M_2$ is the CREF of $M$.

For the example you gave, this tells us that
$$
M = \left(\begin{array}{ccc} 1 & 0 & -1\\ 2 & 0 & -2\\ 0 & 1 & 3 \end{array}\right)
$$
has the desired RREF and CREF.
Proof: To prove that the RREF of $M$ is equal to $M_1$, it suffices to show that $\ker(M) = \ker(M_1)$ (where $\ker(M)$ denotes the kernel AKA nullspace of $M$).  It automatically holds that $\ker(M_1) \subseteq \ker(M_2M_1)$, so it suffices to show the reverse inclusion.
Partition off any zero rows of $M_1$ and the zero columns of $M_2$. That is, write
$$
M_1 = \pmatrix{R_1\\0}, \quad M_2 = \pmatrix{R_2 & 0},
$$
where $R_1$ has no zero rows and $R_2$ has no zero columns. Note the the rows of $R_1$ are linearly independent and the columns of $R_2$ are linearly independent.
With block-matrix multiplication, note that
$$
M_2M_1 = \pmatrix{R_2 & 0} \pmatrix{R_1\\0} = R_2R_1.
$$
Because the columns of $R_2$ are linearly independent, we note that $R_2 x = 0 \implies x = 0$. Thus, we have
$$
M_2M_1 x = 0 \implies R_2(R_1 x) = 0 \implies R_1 x = 0 \implies M_1 x = 0,
$$
so we indeed have $\ker(M_2M_1) \subseteq \ker(M_1)$, which was what we wanted.
Similarly, to prove that the CREF of $M$ is equal to $M_2$, it suffices to show that $\ker(M^T) = \ker(M_2^T)$. However, we have
$$
M^T = (M_2M_1)^T = M_1^T M_2^T.
$$
Because $M_1^T$ is in CREF and $M_2^T$ is in RREF, the desired result from our previous work.

Note that in general, it is possible for many matrices to have the same RREF and the same CREF.  For example: any matrix of the form
$$
\pmatrix{a_{11} & a_{12} & 0\\ a_{21} & a_{22} & 0\\0 & 0 & 0}
$$
with $a_{11}a_{22} \neq a_{12}a_{21}$ will have RREF and CREF equal to
$$
\pmatrix{1&0&0\\0&1&0\\0&0&0}.
$$
If we define $R_1$ and $R_2$ as above and let $r$ denote the rank of $M$, then for any $r \times r$ invertible matrix $S$, the matrix
$$
R_2 S R_1
$$
will have RREF $M_1$ and CREF $M_2$. In fact, every matrix with this RREF and CREF can be expressed in the above form for some choice of $S$. This can be shown, for instance, with the help of the singular value decomposition.
