The short answer is: Yes, the formula you showed generates all primitive Pythagorean triples and all primitive triples will have this form.$\space\space$
(Primitives have $\space(q,p) \space$ of opposite parity and $\space GCD(q,p)=1.\quad$
It also generates trivial triples where one-or-more of $\space (x,y,z) \space$ is zero, and imprimitives where
$\space GCD(x,y,z)>1.\quad$
Let's begin with this arbitrary expression of Euclid's formula:
$ A=m^2-k^2,\quad B=2mk,\quad C=m^2+k^2.\quad$
We can see
$$\space m=0\implies A<0\space\land\space B=0 \\ \veebar\\ k=0\implies\space A=C\space\land\space B=0$$
$$\\$$
$$m<0\space\veebar k<0\space\implies B<0$$
$$\\$$
$$GCD(m,k)>1 \implies GCD(A,B,C)>1$$
If we vary Euclid's formula slightly, we get
\begin{align*}
A&=(2n-1+k)^2-k^2\quad= &(2n-1)^2+& 2(2n-1)k \\
B&=2(2n-1+k)k \qquad\quad=& & 2(2n-1)k+ 2k^2\\
C&=(2n-1+k)^2+k^2 \quad =&(2n-1)^2+& 2(2n-1)k+ 2k^2
\end{align*}
and this formula produces a non-trivial Pythagoreaan triple for ever pair of natural numbers $\space (n,k).\quad$ It produces about a third of the imprimitive triples as Euclid's formula but it does produce "some" because
$\space GCD(A,B,C)=(2x-1)^2,\space x\in\mathbb{N}.\quad$ Note:
$\space GCD(A,B,C)=(2(1)-1)^2=1\space$ includes all primitives.
We can see the $\space Sets\space$ of triples it produces in the table below. I actually developed the formula in 2009 by arranging hundreds of the smallest triple into this periodic-like table and noticing $\quad C-B=(2n-1)^2\space $ and
$\quad A_{n+1} - A_n=2(2n-1).\quad$
$$\begin{array}{c|c|c|c|c|c|}
Set_n & k=1 & k=2 & k=3 & k=4 &\cdots \\ \hline
Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41 &\cdots\\ \hline
Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 &\cdots\\ \hline
Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 &\cdots\\ \hline
Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 &\cdots\\ \hline
\end{array}$$
We can see
$\space C-B=(2n-1)^2\space$ and
$\space n=1\implies C-B=1\implies$ all members of $\space Set_1\space$ are primitive. $\space\space$
If $\space (2n-1)\space$ is prime, then this variation will repeatedly "generate $(2n-1)-1$ primitives and skip an imprimitive."
\begin{align*}
&A=(2n-1)^2+&2(2n-1)\bigg(k+\bigg\lfloor\frac{(k-1)}{(2n-2)}\bigg\rfloor\bigg)&\qquad\\
&B=&2(2n-1)\bigg(k+\bigg\lfloor\frac{(k-1)}{(2n-2)}\bigg\rfloor\bigg)&\qquad+2\bigg(k+\bigg\lfloor\frac{(k-1)}{(2n-2)}\bigg\rfloor\bigg)^2\\
&C=(2n-1)^2+&2(2n-1)\bigg(k+\bigg\lfloor\frac{(k-1)}{(2n-2)}\bigg\rfloor\bigg)&\qquad+2\bigg(k+\bigg\lfloor\frac{(k-1)}{(2n-2)}\bigg\rfloor\bigg)^2
\end{align*}
If $\space (2n-1)\space$ is composite, then
$\space(A,B,C)\space$ is imprimitive when
$\space GCD(n,k)>1.\quad$
Could this new formula be missing triples in the subset where $\space GCD(A,B,C)\space$ is and odd square? Let us assume that some other $A$-interval $(x)$ will include $A$-values in an extended version of the Table and reveal missing triples. If we derive $\space (B,\space C)\space$ from $\space A\space$ using the Pythagorean theorem, we get:
$$A=(2n-1)^2+xk \quad\implies\\
B=xk+\frac{x^2k^2}{2(2n-1)^2 }\\
\land
\\
C=(2n-1)^2+xk+\frac{x^2k^2}{2(2n-1)^2}$$
The $B$ and $C$ functions will not yield integers if
$\space 2(2n-1)^2\not\mid x^2k^2\space$ for most values of $k$ so that $F(n,k)$ is not missing any triples (especially primitives) in the subset where $\space GCD(A,B,C)\space$ is an odd square –– and $F(n,k)$ is a subset of $F(m,k)$.
$\quad\therefore :\quad$All primitives are represented by both formulas.
