Assuming that when you say "two positive solutions" you mean "two positive distinct solutions" you are correct that you need the discriminant to be positive, and you've correctly computed this.
But I don't think you figured out the correct condition for the roots to be positive. You can have a quadratic with vertex in the positive numbers, but one root positive and one root negative. For example, take $(x+1)(x-4)=x^2-3x-4$. The minimum occurs at $x=\frac{3}{2}$, but one root is negative.
So while having the vertex in the positive numbers is necessary, it is not sufficient.
Instead, think about the quadratic formulas for the roots. Since the roots are $\frac{(m+1) \pm \sqrt{D}}{2}.$
The smaller root will occur when we take $\frac{m+1-\sqrt{D}}{2}$, so you want $m+1-\sqrt{D}\gt 0$, which means you want $m+1\gt \sqrt{D}$.
Edit and fix. If you square both sides, we get $(m+1)^2 \gt D = (m+1)^2 - 4(m+4)$, which means you want $0\gt -4(m+4)$, which means you want $-4(m+4)\lt 0$, or $m+4 \gt 0$. But this may add spurious solutions (because of the squaring), which occur when $m+1 \lt - \sqrt{D}\lt 0$. So we need to take out these possibilities. These spurious solutions are introduced example when $|m+1|\geq \sqrt{D}$ and $m+1\lt 0$; that is, $(m+1)^2\geq D$ and $m+1\lt 0$. As above, these occur when $m+4\gt 0$ and $m+1\lt 0$, that is, when $-4\lt m\lt -1$.
So we need $m\gt -4$, and not to have $m\lt -1$. This gives $m\geq -1$ in summary.
So the conditions are: $m\in (-\infty,-3)$ and $m\geq -1$ (impossible); or $m\in (5,\infty)$ and $m\geq -1$, which yield $m\in (5,\infty)$.
If you are okay with a single double root which is positive, then you also allow $D=0$, which means you need to include $x=5$ into your solution set.