= Solution
Use the <strong form of Hensel lemma>: for $f\in\mathbb Z_p[T]$ and $a\in\mathbb Z_p$, if
$$
v_p(f(a))>2v_p(f'(a)),
$$
there is a unique <polynomial root> $b$ in the ball $v_p(b-a)>v_p(f'(a))$. In particular, a simple <polynomial root> modulo $p$ lifts uniquely in its residue class. Newton iteration gives $v_p(b-a)=v_p(f(a))-v_p(f'(a))$ when $f(a)\ne0$.
For odd $p\notin\{5,7,13\}$, the sets
$$
\{5x^2:x\in\mathbb F_p\},
\qquad
\{-13-7y^2:y\in\mathbb F_p\}
$$
each have $(p+1)/2$ elements, so they intersect. This gives a zero modulo $p$ with $z=1$. The derivative with respect to $z$ is $26$, nonzero modulo $p$, so <Hensel's lemma> lifts $z$ while holding $x,y$ fixed. This is the elementary proof of <isotropy of nondegenerate ternary quadratic forms over finite fields> followed by a simple-root lift.
At $p=5$, the residue vector $(0,1,1)$ is a zero, and the derivative $14y$ is nonzero. At $p=13$, use $(3,1,0)$, for which the value is 52 and the derivative $10x=30$ is nonzero modulo 13. Again a single-variable Hensel lift suffices.
At $p=2$, hold $x=2,z=1$ and use $f(Y)=7Y^2+33$. At $Y=1$, $f(1)=40$ has <valuation> 3, while $f'(1)=14$ has <valuation> 1. Since $3>2$, the strong Hensel inequality gives a <polynomial root> $y\in\mathbb Z_2$, indeed $y\equiv1\pmod4$. This yields a nonzero solution.
At $p=7$, suppose a nonzero solution exists and scale it so that all coordinates are in $\mathbb Z_7$ and at least one is a <unit>. Reduction modulo 7 gives $z^2=5x^2$. Since 5 is not a square modulo 7, both $x,z$ are divisible by 7. In the original equation their terms are then divisible by 49, so $7y^2$ is divisible by 49 as well, forcing $y$ divisible by 7. This contradicts the normalization. Thus <local isotropy of the five seven thirteen form> gives
$$
\boxed{\text{The unique obstructing prime is }p=7.}
$$
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