I am not contributing something new. I am taking a result given in Watson and arriving at the same condition that
@renormalize arrived at.
On page 148 of Watson is the formula:
\begin{align*}
J_\mu (az) J_\nu (bz) & = \dfrac{ (\frac{1}{2} az)^\mu (\frac{1}{2} b z)^\nu}{\Gamma (\nu+1)}
\nonumber \\
& \times \sum_{m=0}^\infty \dfrac{(-1)^m (\frac{1}{2} az)^{2m} \;_2F_1 \left( -m , - \mu - m ; \nu+1 ; b^2/a^2 \right)}{ m! \Gamma (\mu+m+1)}
\end{align*}
Here
https://books.google.co.uk/books?id=qy1GNv2ovHQC&lpg=PA59&dq=integral+product+bessel+functions&pg=PA59&redir_esc=y#v=onepage&q=integral product bessel functions&f=false they give a slightly different version of the formula:
\begin{align*}
J_\mu (az) J_\nu (bz) & = \dfrac{ (\frac{1}{2} az)^\mu (\frac{1}{2} b z)^\nu}{\Gamma (\mu+1)}
\nonumber \\
& \times \sum_{m=0}^\infty \dfrac{(-1)^m (\frac{1}{2} bz)^{2m} \;_2F_1 \left( -m , - \nu - m ; \mu+1 ; a^2/b^2 \right)}{m! \Gamma (\nu+m+1)}
\end{align*}
which is obtained from the Watson formula by simultaneously performing the interchanges ##a \leftrightarrow b## and ##\mu \leftrightarrow \nu##. We'll use the second version. From which we have:
\begin{align*}
& \dfrac{J_{n+1} ((n+1)x)}{n+1} \dfrac{J_{n+2} ((n+2)x)}{n+2}
\nonumber \\
& = \dfrac{ ((n+1)x/2)^{n+1} ((n+2)x/2)^{n+2}}{(n+1) (n+2) (n+1)!} \sum_{m=0}^\infty \dfrac{(-1)^m ((n+2) x/2)^{2m}}{ m! (n+m+2)! }
\nonumber \\
& \qquad \qquad \times \;_2F_1 \left( -m , - n - 2 - m ; n+2 ; \left( \frac{n+1}{n+2} \right)^2 \right)
\end{align*}
or
\begin{align*}
& \dfrac{J_{n+1} ((n+1)x)}{n+1} \dfrac{J_{n+2} ((n+2)x)}{n+2}
\nonumber \\
& = \dfrac{(n+1)^n (n+2)^{n+1}}{(n+1)!} \sum_{m=0}^\infty \dfrac{(-1)^m (n+2)^{2m}}{ m! (n+m+2)!}
\nonumber \\
& \qquad \qquad \times \;_2F_1 \left( -m , - n - 2 - m ; n+2 ; \left( \frac{n+1}{n+2} \right)^2 \right) \left( \frac{x}{2} \right)^{2n+2m+3}
\end{align*}
Then,
\begin{align*}
x \phi (x) & = \sum_{n=0}^\infty \dfrac{J_{n+1} ((n+1)x)}{n+1} \dfrac{J_{n+2} ((n+2)x)}{n+2}
\nonumber \\
& = \sum_{n=0}^\infty \sum_{m=0}^\infty \dfrac{(-1)^m (n+1)^n (n+2)^{2m+n+1}}{(n+1)! m! (n+m+2)!}
\nonumber \\
& \qquad \qquad \times \;_2F_1 \left( -m , - n - 2 - m ; n+2 ; \left( \frac{n+1}{n+2} \right)^2 \right) \left( \frac{x}{2} \right)^{2n+2m+3}
\nonumber \\
& = \sum_{n=0}^\infty \sum_{m=0}^\infty s_{n,m} \left( \frac{x}{2} \right)^{2n+2m+3}
\end{align*}
where ##s_{n,m}## is defined by
\begin{align*}
s_{n,m} & = \dfrac{(-1)^m (n+1)^n (n+2)^{2m+n+1}}{(n+1)! m! (n+m+2)!}
\;_2F_1 \left( -m , - n - 2 - m ; n+2 ; \left( \frac{n+1}{n+2} \right)^2 \right)
\end{align*}
The double sum can be written
\begin{align*}
\sum_{n=0}^\infty \sum_{m=0}^\infty s_{n,m} \left( \frac{x}{2} \right)^{2n+2m+3} = \sum_{m=0}^\infty \sum_{n=0}^m s_{n,m-n} \left( \frac{x}{2} \right)^{2m+3}
\end{align*}
Here
\begin{align*}
s_{n,m-n} & = \dfrac{(-1)^{m-n} (n+1)^n (n+2)^{2m-n+1}}{(n+1)! (m-n)! (m+2)!}
\;_2F_1 \left( -m + n , - 2 - m ; n+2 ; \left( \frac{n+1}{n+2} \right)^2 \right)
\end{align*}
So
\begin{align*}
\phi (x) = \sum_{m=0}^\infty \sum_{n=0}^m \frac{1}{2} s_{n,m-n} \left( \frac{x}{2} \right)^{2m+2} .
\end{align*}
Writing
\begin{align*}
\phi (x) & = \sum_{m=0}^\infty \sum_{n=0}^m \frac{1}{2} s_{n,m-n} (-1)^m m! (m+2)! \dfrac{(-1)^m}{m! (m+2)!} \left( \frac{x}{2} \right)^{2m+2}
\end{align*}
we see that if ##\phi (x) = J_2 (x)## we would have
\begin{align*}
\sum_{n=0}^m \frac{1}{2} (-1)^m m! (m+2)! s_{n,m-n} = 1 \qquad \text{for } m \geq 0 .
\end{align*}
That is, if ##\phi (x) = J_2 (x)## we would have
\begin{align*}
\sum_{n=0}^m \dfrac{m! (-1)^{-n} (n+1)^n (n+2)^{2m-n+1}
\;_2F_1 \left( n-m , - m - 2 ; n+2 ; \left( \frac{n+1}{n+2} \right)^2 \right)}{2 (n+1)! (m-n)!} = 1
\end{align*}
for ##m \geq 0##. Which is the same condition that
@renormalize arrived at.