Considering this as a quadratic in $y$, it is sufficient (and necessary) that
$n^4 - 4n^2(x^2-z^2)$ is a perfect square.
i.e.
$n^2 - 4(x^2 - z^2)$ is a perfect square, say $q^2$.
Rewriting gives us
$(n-q)(n+q) = 4 (x^2 - z^2)$
If $n=2k$ is even, then $q$ needs to be even too (say $2m$) and
$(k-m)(k+m) = (x-z)(x+z)$
If $n=2k+1$ is odd, then $q$ needs to be odd too (say $2m+1$) and
$(k-m)(k+m+1) = (x-z)(x+z)$
Thus you can pick any $m$, and try to factor the left hand side above (choosing the right one depending on the parity of $n$) into two terms of the same parity.
The equation,
$$wxyz=(w+x+y+z)^2\tag{1}$$
can be solved as a quadratic in $z$,
$$(w + x + y)^2 + \big(2(w + x + y) - w x y\big) z + z^2=0\tag{2}$$
Its discriminant is $wxy(wxy-4\big(w+x+y)\big)$ and must be made a square. It can be shown an initial solution $a,x,y$ can generate an infinite more. Define,
$$axy(axy-4\big(a+x+y)\big)=c^2\tag{3}$$
$$a^2xy(xy-4) =d\tag{4}$$
then a family of solutions to (2) can be given as,
$$w =\frac{a(p-cq)^2}{p^2-dq^2}\tag{5}$$
$$z =-(w+x+y)+\frac{1}{2}\left(wxy\pm\frac{(p-cq)(cp-dq)}{p^2-dq^2}\right)\tag{6}$$
for arbitrary $p,q$. If we want integers, then one can solve the Pell equation $p^2-dq^2=1$. Some examples are,
$$w,x,y = 4,\;4,\;4(p-q)^2;\;\;z=4(p+q)^2\;\text{or}\;\;4(3p-5q)^2,\;\;\text{where}\;p^2-3q^2=1\tag{7}$$
or,
$$w,x,y = 3,\;3,\;6(p-q)^2;\;\;z=6(p+q)^2\;\text{or}\;\;24(p-2q)^2,\;\;\text{where}\;p^2-5q^2=1\tag{8}$$
and so on.
Best Answer
The answer is NO.
The quartic is birationally equivalent to the elliptic curve \begin{equation*} v^2=u(u^2+2u-1) \end{equation*} with \begin{equation*} \frac{x}{y}=\frac{v}{u} \end{equation*}
Pari-GP gives the torsion subgroup of the curve as having only one finite point $(0,0)$, whilst Denis Simon's $\mathbf{ellrank}$ code gives the rank as $0$.
Thus, the only rational point on the curve is $(0,0)$ which does not, clearly, give a solution.