The most common formula for generating Pythagorean triples is Euclid's, shown here as $$A=m^2-k^2\qquad B=2mk\qquad C=m^2+k^2$$ It generates all primitves but also generates trivals, doubles, square multiples, and doubles of square multiples of primitives. A variation of this is
\begin{align*} &A=(2n-1+k)^2-k^2&&=(2n-1)^2+2(2n-1)k\\ &B=2(2n-1+k)k &&=\phantom{(2n-1)^2+{}} 2(2n-1)k+2k^2\\ &C=(2n-1+k)^2+k^2 &&=(2n-1)^2+2(2n-1)k+2k^2 \end{align*} It generates no trivials and all primitives but also generates odd square multiples of primitives.
In the latter, if we let $\,n=1\implies \big(A=2k+1\quad B=2k^2 + 2k\quad C=2 k^2 + 2 k + 1\big),\,$ or $\,k-1\implies \big(A=4n^2-1\quad B=4n\quad C=4n^2+1\big)$ we get only primitives but not all primitives.
Is there a formula that genrates only and all primitives besides the one that generates the ternary tree? I doubt such a formula exists but I would love to see it if it does.
As Will Jagy pointed out in his comment, Euclid's formula satisfies your criteria IF certain conditions are stated. The trick is to find a way to include those conditions in a formula, rather than having them stated independently. Here is as close as I can come; apologies if you think it is falls short of what you seek.
$$r:=2^a(2u-1); s:=2^{a+b}(2v-1); (a,b,u,v) \in \mathbb Z^+$$ $$ m:=\frac{r}{\gcd(r,s)}; k:=\frac{s}{\gcd(r,s)}$$ $$ A=|m^2-k^2|; B=2mk; C=m^2+k^2$$ It is possible to cram all of this into one set of formulas for $(A,B,C)$ in terms of the variables $(a,b,u,v)$ if necessary, but the typography would be horrible.
A limitation similar to $(a,b,u,v) \in \mathbb Z^+$ is (I believe) generally presumed, even if not expressly stated, in all such formulas for Pythagorean triples. The formulas arrived at here for the Pythagorean triples $A,B,C$ meet the criteria of Euler's formula in terms of completeness, ensure that one and only one of $m,k$ is even, and avoid any shared factors among $m,k$, hence among $(A,B,C)$, ensuring primitive triples.