I'm afraid that, even if we assume (as suggested in A.Meyerowitz's comment) that the intention is *integer* solutions of $F(x_1,x_2,x_3,x_4) = 0$, the number of solutions with $\max_i |x_i| \leq B$ will grow at least as $B^2$ as long as there's any nonzero solution, and sometimes the growth will even be a bit faster.

The usual heuristics suggest that if there's no local obstruction (such as there is for W.Zudilim's example of $X_1^2 + X_2^2 + X_3^2 + X_4^2$, with no nontrivial zeros even over $\bf R$) then the number solutions up to height $B$ should grow as $B^2$: there are about $B^4$ candidates, and for each one of them $F(x_1,x_2,x_3,x_4)$ has size at most $B^2$, so if we imagine they're randomly distributed then about $B^4 / B^2 = B^2$ values should be zero.

It is not hard to prove that for some choices of quadratic form $F$, even nonsingular ones, the count is $\gg B^2$. Namely suppose $F(x_1,x_2,x_3,x_4)$ has the form $Q(x_1,x_2) - Q(x_3,x_4)$ for some quadratic $Q$. Then the line $(x_1,x_2,x_1,x_2)$ already gives $B^2$ zeros. Geometrically, the two rulings of the quadric $Q=0$ in ${\bf P}^3$ are defined over ${\bf Q}$, and any line — not just the trivial $\{(x_1:x_2:x_1:x_2)\}$ — will give some positive multiple of $B^2$. Using all the lines in the ruling one finds that in fact the correct growth rate is $B^2 \log B$.

If I remember right it is known that for an unobstructed smooth quadric in ${\bf P}^3$ whose rulings are not rational but Galois conjugate over ${\bf Q}$, the counting function $N_1(F,B)$ is asymptotically proportional to $B^2$ as the heuristics suggested.

This is all consistent with Manin's conjecture: the quadrics with rational rulings are precisely those for which the rational subgroup of the Néron-Severi group has rank 2 rather than 1, which accounts for the extra factor of $\log B$.

rationalbut with if $|x_i|$ means absolute value then indeed the number could well be infinite. So did you mean integer solutions or some height on $\mathbb{Q}$ such as $|\frac{p}{q}|$ being the larger of the absolute values of $p$ and $q$? Of course that would allow more than $B^3$ solutions with $|x_i| \leq B$ $\endgroup$ – Aaron Meyerowitz Jul 2 '11 at 21:36