$x_4=4$ is possible if you break the terms like that but it will not give the non-negative integer solutions. I am mentioning the reason below.

As you have broken terms, you could write $3(x_1 + x_2 + x_3 + x_4) + 4x_4 =22 \implies x_1+x_2+x_3+x_4 = \frac{22-4x_4}{3}$

Since, left side is non-negative integer, right side should also be integer and non-negative which is possible when $x_4=1,4$

When $x_4 = 1$ then our equation becomes $x_1+x_2+x_3+1 = \frac{22 – 4\times 1}{3} \implies x_1 + x_2+x_3 =5$

When $x_4 = 4$ then our equation becomes $x_1+x_2+x_3+4 = \frac{22 – 4\times 4}{3} \implies x_1 + x_2+x_3 =-2$

$ x_1 + x_2+x_3 =-2$ can not produce non-negative integers $x_1,x_2,x_3$ such that their sum will be a negative number.