Prime Plus One

Let $p$ be the prime and $a^2$ be the perfect square. So the problem can be re-written as $p+1=a^2\Rightarrow p=a^2-1\Rightarrow p=(a+1)(a-1).$
Every prime has only two factors,that are $1$ and the number itself. So in the product $(a+1)(a-1)$ one of them should be $1.$ If $(a+1)=1,$ then $a=0$ and $p=0$ , which is not possible since it $p$ is a prime. So, $(a-1)=1\Rightarrow a=2; p=(3)(1)=3.$ Therefore, the perfect square which is $1$ more than a prime is $(p+1)=3+1=4.$ Hence there is only $\boxed1$ perfect square that satisfies the given condition.

Lets say the prime is p and x^2 is the number. Now by the problem x^2=p+1 Then (x-1)(x+1) = p Or
Case 1. x-1=p and x+1=1 or x=0 or p=-1. Not possible

Case 2

X+1=p and x-1=1 hence x=2 or

p=3 (by solving with the method of comparision) Hence there is only 1 prime