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The given equation can be rewritten as

$$3(2y+1)(2z+1)=(2x+1)(2w+1),$$ which suggests that solutions can be generated by first fixing values of $y,z$ and thus $n:=3(2y+1)(2z+1)$, then running $d$ over the divisors of $n$, and setting $x=\frac{d-1}2$ and $w=\frac{n/d-1}2$ .