Let’s assume that the two-digit number $AB$ is greater than or equal to the two-digit number $CD$. The actual values of these numbers are $10A+B$ and $10C+D$, so the difference of their squares is $(10A+B)^2-(10C+D)^2$. If we reverse the digets to get $BA$ and $DC$, the values of these numbers are $10B+A$ and $10D+C$, and the difference of their squares is either $(10B+A)^2-(10D+C)^2$ or $(10D+C)^2-(10B+A)^2$, depending on which is larger.
In the first case, in which $BA$ is at least as large as $DC$, we have $(10A+B)^2-(10C+D)^2=(10B+A)^2-(10D+C)^2\;,$ or, after multiplying out the squares, $\begin{align*}100A^2&+20AB+B^2-100C^2-20CD-D^2\\ &=100B^2+20AB+A^2-100D^2-20CD-C^2\;; \end{align*}$
simplifying this yields the much nicer $99A^2-99B^2=99C^2-99D^2\;,$ or simply $A^2-B^2=C^2-D^2$. It may be helpful to rewrite this as $A^2+D^2=B^2+C^2$. Now you can simply look at the sums of the squares of the digits, as in the following table. I’ve blanked out the cases in which the two digits are identical, and I’ve included only the second one of pairs like $1^2+2^2$ and $2^1+1^2$.
$\begin{array}{r|c} &\color{blue}{0}&\color{blue}{1}&\color{blue}{4}&\color{blue}{9}&\color{blue}{16}&\color{blue}{25}&\color{blue}{36}&\color{blue}{49}&\color{blue}{64}\\ \hline \color{blue}{1}&1\\ \hline \color{blue}{4}&4&5\\ \hline \color{blue}{9}&9&10&13\\ \hline \color{blue}{16}&16&17&20&\color{red}{25}\\ \hline \color{blue}{25}&\color{red}{25}&26&29&34&41\\ \hline \color{blue}{36}&36&37&40&45&52&61\\ \hline \color{blue}{49}&49&50&53&58&\color{red}{65}&74&\color{red}{85}\\ \hline \color{blue}{64}&64&\color{red}{65}&68&73&80&89&100&113\\ \hline \color{blue}{81}&81&82&\color{red}{85}&90&97&106&117&130&145 \end{array}$
The sums that appears in two different places are $\color{red}{25}$, $\color{red}{65}$, and $\color{red}{85}$, corresponding to $\begin{align*}3^2+4^2&=0^2+5^2\;,\tag{1}\\ 4^2+7^2&=1^2+8^2\;,\text{ and}\tag{2}\\ 6^2+7^2&=2^2+9^2\;.\tag{3} \end{align*}$
We’ll take them in order. In $(1)$ the $0$ cannot be $A$ or $C$, since these are supposed to be genuine two-digit numbers. If $B=0$, then $C=5$; in that case $A$ must be $2$ or $3$, and $AB$ is smaller than $CD$, contrary to our original assumption. Thus, $D=0$, $A=5$, and our numbers are either $AB=53$ and $CD=40$, or $AB=54$ and $CD=30$. Both of these pairs work: $53^2-40^2=2809-1600=1209=1225-16=35^2-4^2\;,$ and $54^2-30^2=2916-900=2016=2025-9=45^2-3^2\;.$
In $(2)$ the possibilities are $AB=87$ and $CD=41$, $AB=84$ and $CD=71$, $AB=78$ and $CD=14$, and $AB=48$ and $CD=17$. (Note that the last two are the reversals of the first two; they are still distinct solutions, however.)
Similarly, in $(3)$ the possibilities are $AB=97$ and $CD=62$, $AB=96$ and $CD=72$, and the reversals $AB=79$ and $CD=26$, and $AB=69$ and $CD=27$.
Altogether we have the following ten pairs of numbers as solutions:
$\begin{array}{} (54,30)&(53,40)\\ (87,41)&(84,71)&(78,14)&(48,17)\\ (97,62)&(96,72)&(79,26)&(69,27) \end{array}$
The corresponding differences of squares are:
$\begin{array}{} 2016&1204\\ 5888&2015&5888&2015\\ 5565&4032&5565&4032 \end{array}$
Thus, there are (at least) six answers that Jose could have obtained.
In the second case, in which $BA$ is smaller than $DC$ we have $(10A+B)^2-(10C+D)^2=(10D+C)^2-(10B+A)^2\;;$ if we attempt the same approach, we get
$\begin{align*}100A^2&+20AB+B^2-100C^2-20CD-D^2\\ &=100D^2+20CD+C^2-100B^2-20AB-A^2\;, \end{align*}$
or $101A^2+40AB+101B^2=101C^2+40CD+101D^2\;.$
This time the mixed products don’t disappear, and I see no simple way to investigate possible solutions.