Solve the equation $x! + y!+z!= 2^{v!}$
Assume $x \geq y \geq z$. If $z \geq 3,$ $3|LHS,$ contradiction. Also $2^{v!} = x! + y! + z! \geq 3,$ so $v>1$. Case $1:$ $z=2$ $x! + y! = 2^{v!}-2$. $\mod 3$ gives $y=2$ and so $x! = 2^{v!}-4$. Now $v\geq 3$ and $x!$ is divisible by $4,$ but not by $8,$ which is not possible.
Case $2:$ $z=1$ $x! + y! = 2^{v!}-1$ and thus $y=1$ (otherwise the $LHS$ is even). Now $x! = 2^{v!}-2$ and $x!$ is divisible by $2,$ but not by $4$, so $x<4$. This gives us the solutions $(x, y, z, v) = (2, 1, 1, 2), (3, 1, 1, 3)$ and permutations.
Can anyone correct as much of my test as possible? Does it just work only (2, 1, 1, 2)?
Everything you did is correct, except the only solution, using your $x \ge y \ge z$ ordering, is $(2, 1, 1, 2)$. The solution of $(3, 1, 1, 3)$ you proposed, as stated in the question comments, is not valid. To see this, using your reduced equation of
$$x! = 2^{v!} - 2 \tag{1}\label{eq1A}$$
you get the left side being $3! = 6$, but the right side is $2^{3!} - 2 = 2^6 - 2 = 62$.
As for your comment asking
I believe the key issue is that what you were doing was finding necessary conditions, but they weren't sufficient conditions. In other words, your stated conditions must be satisfied, but there are other additional constraints which aren't satisfied. This is why $(3, 1, 1, 3)$ met your required conditions of $x \lt 4$ with $x!$ having a factor of $2$ but not $4$, although it is not a valid solution.
For more details about this issue of "weird roots", including several related examples, please read Extraneous and missing solutions.