Let $a=3^n$,then $a$ is in exactly $n$ triples(in fact any $a$ if the form $p^n$ satisfayes whrere $p$ is a prime of the form $4k+3$ and a=2^(n+1)).The cruial thing is that these $a^2$ can't be represented as a sum of two squares(for $p=4k+3$ we use a well known fact that if $x$ is a quadratic residue,than $-x$ isn't and for a=2^(n+1) we use $mod4$).

If we let “some positive integer ”be odd$(k)$,since $(a^2-b^2)^2+(2ab)^2=(a^2+b^2)^2$,then $k$ can only be $a^2-b^2$ or $a^2+b^2$. WLOG $k=p_1^{a_1}p_2^{a_2} \cdots p_t^{a_t}$.Moreover we can let $4|k-3$ so $k$ must be $a^2-b^2$,which means the number $i$ such that $4|p_i-3$ is odd.This solution will not use Dirichlet's theorem on arithmetic progressions. Obviously there're infinite prime,so there must have infinite $p=4k+1$ or $p=4k+3$.If there're infinite $p=4k+1$,we let $p_1=3,p_i \equiv 1 \pmod 4 \forall i \ge 2$.If there're infinite $p=4k+3$,since $5,13 \equiv 1 \pmod 4$,the same results is similar to above Since $n=\frac{\prod_{i=1}^{t}(a_i+1)}{2},a_i \in \mathbb{N}^+,$we're done.

We claim that the desired integer is $2^{n+1}$. Assume that we have proved that $2^n$ appears in exactly $n-1$ such triples. Now let's consider $2^{n+1}$. If there is a triple $(x, y, z)=(x, y, 2^t),$ using $\mod 4$ we would obtain an equation of the from $u^2+v^2=1$ with $u, v$ positive integers, a clear contradiction. Now assume that some triple $(x, y, z)$ has $y=2^{n+1}$. If the triple is primitive, then $2^{n+1}=y=2uv, x=u^2-v^2$ for some relatively prime $u, v$ from where it follows $v=1, u=2^n$ and we see that this triple is unique. If the triple is not primitive, we can write $x=2x_1, z=2z_1$ and we get the equation $x_1^2+(2^n)^2=z_1^2,$ but by the inductive hypothesis we have exactly $n-1$ such triples, so the total number for $2^{n+1}$ is exactly $n$. For the base case, note that $2^1$ appears in exactly $0$ triples.

Claim: There are no positive integer solutions to $x^2 + y^2 = 2^{2n}$ Proof: Obvious for $n=1$. Assume true for all $k = n-1$, then for $$x^2 + y^2 = 2^{2n}$$clearly $x,y$ cannot be both odd so $2 \vert x,y$ hence $(\frac{x}{2},\frac{y}{2},2^{n-1})$ is a solution but contradicts the inductive hypothesis and we are done. Claim: There are $n-1$ solutions to $$x^2 + 2^{2n}=y^2$$. Proof: Because $(y-x)(y+x)=2^{2n}$ (note that $x<y$) implies $x \in \{2^k\vert k=1,2,\cdots,n-1 \}$ and we are done. Hence, $2^{n+1}$ has exactly $n$ solutions. $\blacksquare$