The example $x=y=2,z=3$ shows that we can have $K=2$. Now suppose that we have a solution for some $K$. Multiply $x,y,z$ with $p^s$ where $p>2x+3y+3z$ is prime and where $s\ge 1$. Then we have a solution for $(s+1)K$, hence we have a solution for all even $K$. Now suppose that we have a solution where $K$ is odd. Then $x,y,z,2x+3y+3z$ are squares. Hence $2X^2+3Y^2+3Z^2=A^2$. If this has a solution in positive integers, then also a solution with $(X,Y,Z)=1$. Now $2X^2\equiv A^2(3)\implies 3|X,A\implies Y^2+Z^2\equiv 0(3)\implies Y,Z\equiv 0(3)$, contradiction. Hence $K=2,4,6,8,\cdots$

The answer is even positive integers $K$. Claim: $K$ is even Proof: Suppose otherwise. Then note that $x,y,z,2x+3y+3z$ are perfect squares. Choose $x,y,z$ to be positive integers satisfying this but with the smallest possible sum. Firstly note that $x$ and $2x$ are QRs modulo $3$, implying that $3\mid x$. Since $x$ is a square, $9\mid x$. Let $x = 9a$. Now, we have $3\mid 2x + 3y + 3z = 18a + 3y + 3z,$ so $9$ must divide this, meaning $3\mid y + z$. However, $y$ and $z$ must both be QRs modulo $3$, so $3$ divides both $y$ and $z$. Since $y,z$ are squares, we have $9$ divides $y$ and $z$. Let $y = 9b, z = 9c$. Clearly $a, b, c, 2a + 3b + 3c$ are squares. But $a,b,c$ has a smaller sum than $x,y,z$, contradiction. $\square$ We now show that all even $K$ work. Let $K = 2k$ for some integer $k \ge 1$. Consider $x = 7 \cdot 11^{k - 1}, y = 2\cdot 11^{k - 1}, z = 3\cdot 11^{k - 1}$. Note that $\tau(x) = \tau(y) = \tau(z) = 2k = K$ obviously. We have $\tau(2x + 3y + 3z) = \tau(11^{k-1} (2\cdot 7 + 3\cdot 2 + 3\cdot 3) ) = \tau( 29 \cdot 11^{k-1}) = K$, as desired.