The following is an approach that is not by infinite descent, but imitates at the beginning descent approaches to similar problems.
Any square is congruent to $0$, $1$, or $4$ modulo $8$. It follows easily that in any solution of the given equation, $x$, $y$, and $z$ must be even, say $x=2r$, $y=2s$, $z=2t$. Substitute and simplify. We get
Using more or less the same idea, we observe that $r$, $s$, and $t$ must be even. Let $r=2u$, $s=2v$, $t=2w$. Substitute and simplify. We get
Now the descending stops. The right-hand side is congruent to $-1$ modulo $8$, but no sum of $3$ squares can be.
ADDED: I have found a way to make the descent infinite, for proving a stronger result. Look at the equation
We want to show that the only solution is the trivial one $x=y=z=q=0$. The argument is more or less the same as the one above, except that when (after $2$ steps) we reach $-q^2$, we observe that there is a contradiction if $q$ is odd, so now let $q$ be even, and the descent continues. It would probably be more attractive to use $8$ than $16$, and $4q^2$ instead of $16q^2$.