Clearly, it is enough to decide whether their is rational solutions or not (multiply by the square of a common denominator to have integer solutions).
I assume that $a,b,c,d$ are all nonzero (otherwise, you are reduced to known cases).
Multiplying by $a$ and replacing $x$ by $ax$, one may assume WLOG that $a=1$, so we are reduced to the equation $x^2+by^2+cz^2+dz^2=0$. (*)
This step is not necessary, but it allows to write down simpler conditions.
Hasse Minkowski says that (*) has a nonzero rational solution if and only if it has a nonzero solution over $\mathbb{R}$ and $\mathbb{Q}_p$ for all $p$.
Having a solution over the reals is equivalent to say that $b,c,d$ are not all $>0$.
For the $p$-adic case, it depends on the determinant and local Hasse invariants of the rational quadratic form $x^2+by^2+cz^2+dt^2$.
Here, the determinant is the square class of $bcd$, and if $p$ is prime , the local Hasse invariant is $(b,cd)_p(c,d)_p$.
I will below how to define $( , )_p$.
If $r,s$ are two non zero rationals, write $r=p^\alpha u, s=p^\beta v, p\nmid u, p\nmid v$.
Then $(r,s)_p=(-1)^{\alpha\beta\cdot\frac{p-1}{2}}\left(\dfrac{u}{p}\right)^\beta \left(\dfrac{v}{p}\right)^\alpha$, if $p\neq 2$, where $\left(\dfrac{\phantom{a}}{p}\right)$ is the Legendre symbol,
and $\displaystyle (r,s)_2=(-1)^{\frac{u-1}{2}\cdot \frac{v-1}{2}+\alpha\frac{v^2-1}{8}+\beta\frac{u^2-1}{8}}$
Note for later use that $(-1,-1)_p=1$ if $p\neq 2$ and $(-1,-1)_2=-1$.
Note also that if $p\neq 2$, and the $p$-adic valuations of $r,s$ are both zero, $(r,s)_p=1.$
Recall also the following fact.
Fact. Let $r=p^\alpha u, p\nmid u$. Then $r$ is a square in $\mathbb{Q}_p$ if and only if :
If we translate Thm 6 of Chapter IV $\S$ 2 of Serre's "A course in arithmetic", we get that (*) has a nonzero solution over $\mathbb{Q}_p$ if and only if one of the following cases hold:
Case 1. $bcd$ is not a square modulo in $\mathbb{Q}_p^\times$
Case 2. $bcd$ is a square in $\mathbb{Q}_p^\times$ and $(b,cd)_p=(c,d)_p$ if $p\neq 2$
Case 3. $bcd$ is a square in $\mathbb{Q}_2^\times$ and $(b,cd)_2=-(c,d)_2$
All these conditions amounts to computations of finitely many Legendre symbols, because if $p\nmid bcd$ and $p\neq 2$, then either $bcd$ is not a square in $\mathbb{Q}_p^\times $, or $bcd$ is a square in $\mathbb{Q}_p^\times$ but in this case both symbols $(b,cd)_p$ and $(c,d)_p$ are equal to $1$ !!
Hence for odd prime which are not divisors of $bcd$, you are automatically in one of the first two cases.
So you only have to test if you are in one of the three previous cases only for $p\mid 2bcd $. Note that if the $p$-adic valuation of $bcd$ is odd, $bcd$ is automatically not a square, so you can reduce to prime numbers $p$ such that $p\mid 2bcd$ and $v_p(bcd)$ is even !
Hence, you have an algorithm to decide the existence of a nontrivial solution over $\mathbb{Q}$ (note that the Legendre symbol coincide with the Jacobi symbol, so you can compute it without factoring you integers). However, it won't give you an explicit solution.
Example. Consider $3x^2+3\cdot 5y^2+7z^2-2\cdot 7\cdot 23 t^2=0$.
This is equivalent to consider $x^2+3^2 \cdot 5y^2+3\cdot 7z^2-2\cdot 3 \cdot 7\cdot 23 t^2=0$.
Here $b=3^2 \cdot 5, c=3\cdot 7, d=-2\cdot 3\cdot 7 \cdot 23, cd=-2\cdot 3^2\cdot 7^2\cdot 23$ , and $bcd=-2\cdot 3^4\cdot 5\cdot 7^2\cdot 23$.
Since $b,c,d$ are not all positive, we have solutions over the reals.
For the $p$-adic case, we just have to check two cases: $p=3,7$ since they are the only prime divisors of $bcd$ with an even valuation.
For $p=3$, we need to check that $-2\cdot 5\cdot 23$ is a square modulo $3$.
But $-2\cdot 5\cdot 23=1 \ [3]$, which is a square.
Now since $( r, s)_p$ only depends on the square classes of $r$ and $s$, we have
$(b,cd)_3= (5,-2\cdot 23)_3=(-1)^{\frac{5-1}{2}\frac{-46-1}{2}}=1$ and
$(c,d)_3=(3\cdot 7,-2\cdot 3\cdot 7\cdot 23)_3 =(-1)^{\frac{3-1}{2}}\left(\dfrac{7}{3}\right) \left(\dfrac{-2\cdot 7\cdot 23}{3}\right)=-1\cdot 1\cdot -1=1.$
Hence we are in Case 2.
For $p=7$, we need to check that $-2\cdot 5\cdot 23$ is a square modulo $7$.
But $-2\cdot 5\cdot 23=1 \ [7]$, which is a square.
Since $7\nmid 5,2$ and $3$, $(b,cd)_7=1$. Now $(c,d)_7=(3\cdot 7,-2\cdot 3\cdot 7\cdot 23)_7=(-1)^{\frac{7-1}{2}}\left(\dfrac{3}{7}\right) \left(\dfrac{-2\cdot 3\cdot 23}{7}\right)=-1\cdot -1\cdot 1=1.$
Hence we are in Case 2.
All in all, the original equation must have a non trivial solution.
This is indeed the case since $3\cdot 2^2+3\cdot 5\cdot 3^2+7\cdot 5^2-2\cdot 7\cdot 23\cdot 1^2=0$.