According to Mathematica,
$ _2F_1\left(\frac{v+2}{2},\frac{v+3}{2};v+1;z\right) = \frac{2^v \left(1+\sqrt{1-z}\right)^{-v} \left(1+v \sqrt{1-z}\right)}{(1+v) (1-z)^{3/2}}. $
I would be surprised if this was not known, especially considering how close it is to the known cases $_2F_1(a,a+1/2;2a+1;z)$ and $_2F_1(a,a+1/2;2a;z)$. Indeed, if $b=(v+2)/2$ and $c = (v-1)/2$ then
$\begin{align} _2F_1\left(\frac{v+2}{2},\frac{v+3}{2};v+1;z\right) &= {}_2F_1\left(b,b+\frac{1}{2};2b-1;z\right) \\ &= (1-z)^{-3/2} {}_2F_1\left(c,c+\frac{1}{2};2c+2;z\right). \end{align}$
The DLMF gives Prudnikov et al. (1990, pp. 468ā488) as a reference for elementary representations, which I'll check out from the library tomorrow.
Edit.
With a little coaxing, Mathematica seems able to evaluate $ _2F_1\left(\frac{v+u}{2},\frac{v+u+1}{2};v+1;z\right) $
for all nonnegative integer $u$ through the use of the integral representation
$ _2F_1(a,b;c;z) = \frac{\Gamma(c)}{\Gamma(b)\,\Gamma(c-b)} \int_0^1 \frac{t^{b-1} (1-t)^{c-b-1}}{(1-t z)^a}\,dt. $
For even $u$ we can evaluate it using the code
u=2 n; a=(v+u)/2; b=(v+u+1)/2; c=v+1; Expand[Integrate[t^(b-1) (1-t)^(c-b-1)/(1-t z)^a,{t,0,1}, Assumptions->Re[v]>u-1&&Re[z]<=1] Gamma[c]/(Gamma[b] Gamma[c-b])]//FullSimplify
One such example is
$\begin{align} &_2F_1\left(\frac{v+4}{2},\frac{v+5}{2};v+1;z\right) \\ &\qquad= \frac{2^v v^2 \left(1+\sqrt{1-z}\right)^{-v} \Gamma(v+1) \left(v (1-z)^{3/2}-6 (z-1)\right)}{(1-z)^{7/2} \Gamma(v+4)} \\ &\hspace{2cm} + \frac{2^v \left(1+\sqrt{1-z}\right)^{-v} \Gamma(v+1) \left(6 + 9 z+v \sqrt{1-z} (11+4 z)\right)}{(1-z)^{7/2} \Gamma(v+4)}. \end{align}$
For odd $u$ we can reduce it back to the even case (sort of) by using the identity
$\begin{align} &{}_2F_1\left(\frac{v+u}{2},\frac{v+u+1}{2};v+1;z\right) \\ &\qquad = \frac{2 v {}_2F_1\left(\frac{v+u-1}{2},\frac{v+u}{2},v,z\right)-(v-u+1) {}_2F_1\left(\frac{v+u-1}{2},\frac{v+u}{2},v+1,z\right)}{v+u-1}, \end{align}$
which is courtesy of Mathematica and is probably some combination of Gauss' relations for the contiguous hypergeometric functions.
The code is thus
u=2 n+1; a=(v+u-1)/2; b=(v+u)/2; c1=v; c2=v+1; Expand[ (2 v Integrate[t^(b-1) (1-t)^(c1-b-1)/(1-t z)^a,{t,0,1}, Assumptions->Re[v]>u+1&&Re[z]<=1] Gamma[c1]/(Gamma[b] Gamma[c1-b]) -(v-u+1) Integrate[t^(b-1) (1-t)^(c2-b-1)/(1-t z)^a,{t,0,1}, Assumptions->Re[v]>u-1&&Re[z]<=1] Gamma[c2]/(Gamma[b] Gamma[c2-b]))/(v+u-1)]//FullSimplify
And an example,
$\begin{align} &{}_2F_1\left(\frac{v+3}{2},\frac{v+4}{2};v+1;z\right) \\ &\qquad = \frac{2^v v \left(1+\sqrt{1-z}\right)^{-v} \Gamma(v+1) \left(3\sqrt{1-z} + v(1-z)\right)}{(1-z)^{5/2} \Gamma(v+3)} \\ &\hspace{2cm} + \frac{2^v \left(1+\sqrt{1-z}\right)^{-v} \Gamma(v+1) \left(2+z\right)}{(1-z)^{5/2} \Gamma(v+3)}. \end{align}$