Is it possible for $x+y+z, xy+yz+zx$, and $xyz$ to be perfect squares at the same time for positive integer values of $x,y,z$?

Yes. By straightforward search the smallest example is $\lbrace x,y,z \rbrace = \lbrace 45,64,180 \rbrace$, with $$ (t+45) (t+64) (t+180) = t^3 + 17^2 t^2 + 150^2 t + 720^2. $$ Given any solution $(x,y,z)$ we may produce infinitely many others (other than the trivial scaling $(c^2 x, c^2 y, c^2 z)$) by using the theory of elliptic curves to find rational $z'$ such that $x+y+z'$, $xy+yz'+z'x$, and $xyz'$ are all squares, at which point $(d^2 x, d^2 y, d^2 z')$ works for any integer $d>0$ such that $d^2 z' \in {\bf Z}$. For example, in $\lbrace 45,64,180 \rbrace$ we may replace $64$ by $(460163992/28591599)^2$, and then multiply through by $28591599^2$ to obtain the new solution $$ \lbrace 28591599^2 \cdot 45, \phantom{+} 460163992^2, \phantom{+} 28591599^2 \cdot 180 \rbrace . $$ A complete parametrization is not possible, because it would be tantamount to a rational parametrization of the surface $$ S: xy + yz + zx = r^2, \phantom{and} (x+y+z)xyz = s^2 $$ in projective $(1:1:1:1:2)$ space, and that surface is K3. If I did this right, $S$ is a "singular" K3 surface, i.e. has Picard number $20$ which is maximal for a K3 surface in characteristic zero, and the NéronSeveri group ${\rm NS}(S)$ has rank $20$ and discriminant $48$, and consists of (classes of) divisors defined over ${\bf Q}(i)$. It is actually quite common for natural Diophantine equations to give rise to K3 surfaces of maximal or nearlymaximal Picard number, but that's a story for another time. 

