A possible way to solve this problem is to use the classical Diophantus formulas for the generation of triples of the form $x^2+y^2=z^2$. As shown below, this method leads to a final probability of $\frac {662}{35910} \approx 1.8 \%$. I checked this result by a direct counting algorithm and got the same value.
According to Diophantus parametrization, triples of the form $x^2+y^2=z^2$ with $x,y,z $ integers can be generated by choosing a pair of integers $(a,b)$ (with $a>b$) and setting $x=a^2-b^2$, $y=2ab$, so that $z=a^2+b^2$. In particular, if $a $ and $b $ are coprime and of opposte parity, we get a "primitive" triple (i.e. a triple where $gcd (x,y,z)=1$). Interestingly, all primitive triples can be generated in this way. For example, the pair $(2,1)$ generates $(3,4,5) $, the pair $(3,2)$ generates $(5,12,13) $, and so on. Also, when a primitive triple $(x,y,z) $ is identified, we can generate infinite other triples by simply multiplying it to an integer $n $, that is to say writing $(nx,ny,nz) $. All triples where a sum of two squared integers is a perfect square can be generated in this way.
Coming back to our random choice of two integers $x $ and $y $ in the range between $0$ and $189$, we can firstly note that there are $189+189=378$ trivial cases of the form $(0,y) $ or $(x,0) $ - i.e. cases where we pick $0$ in one of the two trials - that satisfy the condition that $x^2+y^2$ is a perfect square.
Focusing on the cases where $x $ and $y $ are both different from zero, we can count the number of cases where $x^2+y^2$ is a perfect square by identifying the Diophantus pairs $(a,b) $ (with $a >b $) that generate primitive triples (i.e. with $a $ and $b $ coprime and of opposte parity) and that satisfy the inequalities $x=a^2-b^2 \leq 189$ and $y=2ab \leq 189$. Then, we have to count - for each of these primitive triples - how many triples of the form $(nx,ny,nz) $ satisfy the same inequalities. Noting that the curves $a^2-b^2 = 189$ and $2ab =189$ cross at $a=\sqrt{189 (1+\sqrt{2})/2} \approx 15.1 $, we get that the maximal theoretical value of $a $ is $15$ (for higher integer values, at least one inequality is not satisfied). We can also observe that, for $a=15$, the only value of $b $ that satisfies the inequalities is $b=6$, but since the pair $(15,6) $ violates the assumption that $a $ and $b $ must be coprime, we can restrict our analysis to the values of $a \leq 14$. In other words, we have to calculate
$$\sum_{a,b} \left \lfloor \frac {189}{max (2ab,a^2-b^2)} \right \rfloor$$
where $max $ indicates the higher between the two quantities representing $x $ and $y $, and the summation extends over all pairs $(a,b) $ that satisfy the conditions described above, in the range $1 \leq a \leq 14$.
We can now identify these pairs of integers $(a,b) $, and count how many triples are generated from each pair. It is not difficult to show that, for $a \leq 10$, the two inequalities are satisfied for all values of $b < a$. So we can include all $b <a $ where the two integers are coprime and of opposite parity. Thus our counting procedure begins as follows:
$$(2,1) \rightarrow \lfloor 189/4 \rfloor=47$$
$$(3,2) \rightarrow \lfloor 189/12 \rfloor=15$$
$$(4,1) \rightarrow \lfloor 189/15 \rfloor=12$$
$$(4,3) \rightarrow \lfloor 189/24 \rfloor=7$$
$$(5,2) \rightarrow \lfloor 189/21 \rfloor=9$$
$$(5,4) \rightarrow \lfloor 189/40 \rfloor=4$$
$$(6,1) \rightarrow \lfloor 189/35 \rfloor=5$$
$$(6,5) \rightarrow \lfloor 189/60 \rfloor=3$$
$$(7,2) \rightarrow \lfloor 189/45 \rfloor=4$$
$$(7,4) \rightarrow \lfloor 189/56 \rfloor=3$$
$$(7,6) \rightarrow \lfloor 189/84 \rfloor=2$$
$$(8,1) \rightarrow \lfloor 189/63 \rfloor=3$$
$$(8,3) \rightarrow \lfloor 189/55 \rfloor=3$$
$$(8,5) \rightarrow \lfloor 189/80 \rfloor=2$$
$$(8,7) \rightarrow \lfloor 189/112 \rfloor=1$$
$$(9,2) \rightarrow \lfloor 189/77 \rfloor=2$$
$$(9,4) \rightarrow \rfloor 189/72 \rfloor=2$$
$$(9,8) \rightarrow \lfloor 189/144 \rfloor=1$$
$$(10,1) \rightarrow \lfloor 189/99 \rfloor =1$$
$$(10,3) \rightarrow \lfloor 189/91 \rfloor=2$$
$$(10,7) \rightarrow \lfloor 189/140 \rfloor = 1$$
$$(10,9) \rightarrow \lfloor 189/180 \rfloor = 1$$
For $a=11$ to $15$ some pairs $(a,b)$ with $a $, $b $ coprime and of opposite parity must be rejected because do not satisfy the two inequalities (for example, $(11,10) $, which gives $2ab=220 >189$). Then, continuing our counting with the same procedure, we get:
$$(11,2) \rightarrow 1$$
$$(11,4) \rightarrow 1$$
$$(11,6) \rightarrow 1$$
$$(11,8) \rightarrow 1$$
$$(12,1) \rightarrow 1$$
$$(12,5) \rightarrow 1$$
$$(12,7) \rightarrow 1$$
$$(13,2) \rightarrow 1$$
$$(13,4) \rightarrow 1$$
$$(13,6) \rightarrow 1$$
$$(14,3) \rightarrow 1$$
$$(14,5) \rightarrow 1$$
The total count, summing the numbers above, is $142$. This expresses the number of valid triples $x^2+y^2=z^2$ where $x $ is odd and $y $ is even and both are different from zero. Due to the symmetry of the problem, we have to count other $142$ triples where $x $ and $y $ are switched (e.g., $(3,4,5)$ and $(4,3,5) $). This gives a total of $284$ valid triples where $x $ and $y $ are different from zero.
Adding to these the $378$ triples where $x $ or $y $ is zero, we get a total number of $284+378=662$ valid triples that satisfy the conditions given by the OP. This corresponds to the number of choices of two random numbers $x $ and $y $ in the range from $0$ to $189$, so that $x^2+y^2$ is a perfect square.
Because the total possible choices of $x $ and $y $, if picked without replacement, are $190 \cdot 189=35910$, we conclude that the probability that $x^2+y^2$ is a perfect square is $662/35910 \approx 1.8 \%$.