Of course, there is no general formula of the type you wanted
but a whole bunch of the formulae expressing the product of
two $_2F_1$ by hypergeometric (or nearly hypergeometric) means.
They are known as Orr-type theorems and can be found in
Slater's book "Generalized hypergeometric functions", Section 2.5
(there are some instances in Bailey's "Generalized hypergeometric series"
as well). A famous example of this type is Clausen's identity
$$
{}_2F_1(a,b;a+b+\tfrac12\mid z)^2
={}_3F_2(2a,2b,a+b;2a+2b,a+b+\tfrac12\mid z).
$$
In addition, you can use the contiguous relations
[Slater, Section 1.4] which allow one to produce linear
relations between any three functions of the form
${}_2F_1(a+n,b+m;c+k\mid z)$ where $n,m,k\in\mathbb Z$,
as well as the transformation [Slater, Section 1.7.1]
$$
{}_2F_1(a,b;c\mid z)
=(1-z)^{-a}{}_2F_1\biggl(a,c-b;c\Bigm|\frac{z}{z-1}\biggr).
$$

I do not see however that something spectacular happens for
your particular product. In fact, the algorithm described
in the (already mentioned) book "$A=B$" decides whether the
expression $a_k$ given by
$$
\sum_{k=0}^\infty a_kz^k
:={}_2F_1(-n,-n+1;2\mid z){}_2F_1(-n-1,-n+3;2\mid z)
$$
(so that each $a_k$ is a single hypergeometric sum) can be
represented as a sum of *finite* rational terms. If this is
the case (which I really doubt), then you will have your wanted
product as a finite sum of hypergeometric functions; if not,
then this is the *proof* that you have no expression of this type.