Somani, using ChatGPT, has given a negative answer. In fact, for any $a\geq 2$, if $c=8a^2+8a+1$,\[\binom{2a}{a}\binom{4a+4}{2a+2}\binom{2c}{c}= \binom{2a+2}{a+1}\binom{4a}{2a}\binom{2c+2}{c+1}.\]Further families of solutions are given in the comments by SharkyKesa.
This was earlier asked about in a
MathOverflow question, in response to which Elkies also gave an alternative construction which produces solutions - at the moment it is not clear whether Elkies' argument gives infinitely many solutions (although I believe it can).
