= Solution
For the uniform <prior>, write $s=x+y$ for the pooled success count in $m+n$ trials. Conditional on $s$, exchangeability makes all placements of the successes equally likely. The successes in the second batch therefore have a <hypergeometric distribution>, and the factor in question is
$$
B=\frac{\binom{s}{y}\binom{m+n-s}{n-y}}{\binom{m+n}{n}}
=\Pr(Y=y\mid X+Y=s)\quad\text{at }s=x+y.
$$
This is <hypergeometric allocation of exchangeable Bernoulli counts>: sample $n$ positions without replacement from a population of $m+n$ positions containing $s$ successes. The <uniform-count predictive property> gives $\Pr(X+Y=s)=1/(m+n+1)$ and $\Pr(X=x)=1/(m+1)$, so
$$
\boxed{\Pr(Y=y\mid X=x)=\frac{m+1}{m+n+1}\,B.}
$$
As $y$ varies with $x$ fixed, $s=x+y$ also varies. Therefore $B$ alone is not a normalized <hypergeometric distribution> over $y$; its normalization factor is essential.
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