Is it possible to simplify the product of two modified Bessel functions of the first kind with order zero, \begin{align} \Bigg( \sum_{n=0}^{\infty} \dfrac{ \Big( \frac{x^{2}}{4} \Big)^{n} }{n! \cdot n!} \Bigg) \cdot \Bigg( \sum_{k=0}^{\infty} \dfrac{ \Big( \frac{x^{2}}{4} \Big)^{k} }{k! \cdot k!} \Bigg) \text{?} \end{align}
I have gotten as far as \begin{align} & \Bigg( \sum_{n=0}^{\infty} \dfrac{ \Big( \frac{x^{2}}{4} \Big)^{n} }{n! \cdot n!} \Bigg) \cdot \Bigg( \sum_{k=0}^{\infty} \dfrac{ \Big( \frac{x^{2}}{4} \Big)^{k} }{k! \cdot k!} \Bigg) \\ &= \sum_{n=0}^{\infty} \Big( \frac{x^{2}}{4} \Big)^{n} \sum_{k=0}^{n} \frac{1}{k! \cdot k!} \cdot \frac{1}{(n-k)! \cdot (n-k)!} \\ &= \sum_{n=0}^{\infty} \dfrac{ \Big( \frac{x^{2}}{4} \Big)^{n} }{n! \cdot n!} \sum_{k=0}^{n} \binom{n}{k}^{2} \\ &= \sum_{n=0}^{\infty} \binom{2n}{n}^{2} \cdot \dfrac{ \Big( \frac{x}{2} \Big)^{2n}}{(2n)!} \text{;} \end{align} however, I cannot see a way forward. If possible, I'd like to represent the product in terms of another modified Bessel function, beyond the very obvious definition already given above. Perhaps the exponential generating function of the central binomial coefficients could be useful.
Hint: The sequence of squared central binomial coefficients $\left(\binom{2n}{n}^2\right)_{n\geq 0}=(1,4,36,400,4\,900, 63\,504,\ldots)$ is stored as A002894 in OEIS.
Since there is no further exponential generating series given besides $\text{BesselI}(0, 2x)^2=\left(\sum_{n\geq 0}\binom{2n}{n}\frac{x^{2n}}{(2n)!}\right)^2$ we rather do not expect another essentially different representation via Bessel functions.