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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 $\{x_1,x_2,x_1,x_2\}$\{(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$\log B$. 1 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\$.