Answer
See proof below.
Work Step by Step
$g(t)=\sqrt{t}+\sqrt{1+t}-4=t^{1/2}+(1+t)^{1/2}-4$
is continuous and differentiable on $(0,\infty)$.
$g(1)\lt 0, \qquad g(16)\gt 0$
so the Intermediate Value Theorem guarantees that there is a zero between $1$ and $16$.
Check whether $g$ is increasing over $(0,\infty)$.
(If it is, the mentioned zero would be the only zero.)
$g'(t)=\displaystyle \frac{1}{2}(t^{-1/2}-(1+t)^{-1/2})$
$t^{-1/2}=\displaystyle \frac{1}{\sqrt{t}}$ is a positive number, and
$(1+t)^{-1/2}=\displaystyle \frac{1}{\sqrt{t+1}} $has a greater denominator than $\displaystyle \frac{1}{\sqrt{t}},$ so $t^{-1/2}-(1+t)^{-1/2}\gt 0$.
So, for $ t\in (0,\infty),\qquad g'(t)\gt 0$
and $g$ is an increasing function on $(0,\infty)$
So, $g$ has only one zero in $(0,\infty)$