Both of the solutions, x = {-4, -3}, are viable. There are no extraneous solutions.
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An extraneous solution would show up as the x-intercept associated with the missing branch of the square root function. Since both solutions are on the positive-square-root branch, they are both good solutions to the equation.
For graphing purposes, we have subtracted x to get
f(x) = √(3x+13) -5 -x
and we are looking for solutions to f(x)=0.
If you add 5 and square both sides, you get the quadratic equation
3x +13 = (x +5)²
x² +7x +12 = 0
(x +3)(x +4) = 0
x ∈ {-4, -3}