As an addendum of sorts to Mathlover's answer: the differential equation
$(y^\prime)^2=2\left(\frac{y^{n+1}}{n+1}+c_1\right)$
is in fact the very sort of equation that is solved by Abelian (hyperelliptic) functions, in the sense that the integral
$\int\frac{\mathrm dt}{\sqrt{t^{n+1}+c}}$
is a hyperelliptic (Abelian) integral (which, as already noted, can be represented in terms of the Gaussian hypergeometric function), and the original differential equation is solved by the inversion of this integral; i.e., with Abelian functions. Since Mathematica and Wolfram Alpha know nothing about Abelian functions, they are unable to proceed further, and they just leave an implicit equation as a result.
As expected, when $n=2$ or $3$, the hyperelliptic integral becomes an elliptic integral, and thus the differential equation is expected to have elliptic function solutions. I'll carry out the inversion explicitly for those two cases.
For $n=2$, we have, after absorption of arbitrary constants, the expression
$\int\frac{\mathrm dy}{\sqrt{y^3+C_1}}=\sqrt{\frac23}x+C_2$
To make the integral on the left a bit more recognizable, we multiply by a constant on both sides:
$\frac12\int\frac{\mathrm dy}{\sqrt{y^3+C_1}}=\frac12\left(\sqrt{\frac23}x+C_2\right)$
which turns into
$\int\frac{\mathrm dy}{\sqrt{4y^3+4C_1}}=\frac{x}{\sqrt 6}+\frac{C_2}{2}$
and we now recognize the Weierstrass elliptic integral corresponding to the cubic $4y^3-g_2 y-g_3$ on the left side, with the invariants $g_2=0$ and $g_3=-4C_1$. Inversion yields
$y=\wp\left(\frac{x}{\sqrt 6}+\frac{C_2}{2};0,-4C_1\right)$
or, after absorption of arbitrary constants,
$y=\wp\left(\frac{x}{\sqrt 6}+C_2;0,C_1\right)=6\wp\left(x+C_2;0,C_1\right)$
where the homogeneity relation for $\wp$ was used to obtain the final expression.
For $n=3$, we have (changing the form of one of the arbitrary constants for convenience)
$\int\frac{\mathrm dy}{\sqrt{y^4+C_1^4}}=\frac{x}{\sqrt 2}+C_2$
We can try a Weierstrass substitution $y=C_1 \cot\frac{t}{2}$ here:
$\frac1{2C_1}\int\frac{\mathrm dt}{\sqrt{1-\frac12\sin^2 t}}=\frac{x}{\sqrt 2}+C_2$
and we recognize the incomplete elliptic integral of the first kind at this point (and absorbing arbitrary constants while we're at it):
$\frac1{C_1}F\left(2\mathrm{arccot}\frac{y}{C_1}\mid\frac12\right)=\frac{x}{\sqrt 2}+C_2$
We solve for $y$ in stages:
$\begin{align*} F\left(2\mathrm{arccot}\frac{y}{C_1}\mid\frac12\right)&=C_1 x+C_2\\ 2\mathrm{arccot}\frac{y}{C_1}&=\mathrm{am}\left(C_1 x+C_2\mid\frac12\right)\\ y&=C_1\cot\left(\frac12\mathrm{am}\left(C_1 x+C_2\mid\frac12\right)\right)\\ y&=C_1\frac{\sin\left(\mathrm{am}\left(C_1 x+C_2\mid\frac12\right)\right)}{1-\cos\left(\mathrm{am}\left(C_1 x+C_2\mid\frac12\right)\right)}\\ y&=C_1\frac{\mathrm{sn}\left(C_1 x+C_2\mid\frac12\right)}{1-\mathrm{cn}\left(C_1 x+C_2\mid\frac12\right)} \end{align*}$
and that's how Jacobian elliptic functions turn up in the solution.
As it turns out, the hyperelliptic functions for $n$ an odd integer can theoretically be expressed in terms of elliptic functions, but the expressions look rather unwieldy. See Byrd and Friedman for details (p. 252 onwards).