Is there an upper bound on the number of primtive pythagorean triples in which some odd number can be a non-hypoteneuse edge?
Take $15$ for example, we have $15,8,17$ then $15,112,113$ but it's not clear we can keep going forever.
I suspect if infinitely many odd integers were the non-hypoteneuse edge of infinitely many primitive pythagorean triples then we might use the diagonal argument to imply uncountably many different pythagorean triples.
But that would contradict that for $a^2+b^2=c^2$ we have $\left(\dfrac{a}{c}\right)^2+\left(\dfrac{b}{c}\right)^2=1$, putting every triple into bijection with a set of rational points on the unit circle and making them countable. So by that argument there are at most finitely many which are members of infinitely many triples.
A simple way to show there is not is to use the parameterization of primitive Pythagorean triples. Let $m,n$ be coprime and of opposite parity. Then $m^2-n^2,2mn,m^2+n^2$ is a primitive Pythagorean triple. For our use, let $m$ be even and $n=1$, giving the fact that $m^2-1,2m,m^2+1$ is a Pythagorean triple. If you want an odd number that is a leg of $k$ Pythagorean triples we start with $$2^2-1,2\cdot 2,2^2+1\\4^2-1,2\cdot 4,4^2+1\\6^2-1,2\cdot 6,6^2+1\\ \ldots \\(2k)^2-1,2\cdot 2k,(2k)^2+1\\$$ Now multiply each by $(2^2-1)(4^2-1)(6^2-1)\ldots ((2k)^2-1)$ divided by its first element and we will have $k$ triangles all with odd leg $(2^2-1)(4^2-1)(6^2-1)\ldots ((2k)^2-1)$