Twin primes of the form $n^2+1$ and $N^2+3$? Assume that there are infinity many primes of the form  $n^2+1$ and there are infinity many primes of the form  $N^2+3$ , Then could we show that there are infinity primes of the form $n^2+1$  and   $N^2+3$ (twin primes ) ?

Edit: I have edit the question just to show that $n$ and $N$ are not the same
This is certainly not sufficient.  For example, it could be that all the primes of the form $n^2+1$ are of the form $(4k)^2+1$ and all the primes of the form $n^2+3$ are of the form $(4k+2)^2+3$.  Then there would be no twin primes of this type, even if there were infinitely many of each.  Yes, $17$ and $19$ is an example, so I could say that all the primes above $10^{1000}$ satisfy my restriction.  There could then be many twin primes of your type, but not an infinite number.
A (trivial) insight...It's a lot easier to show which primes $n^2+1$ can't produce. Consider: $n^2+1 = n^2-1+2 = (n-1)(n+1)+2$; if $n^2+1 = p$, $p$ prime, then $(n-1)(n+1)+2 = p$, or $(n-1)(n+1) = p - 2$. Clearly $p$ can never be the larger of any two twin prime pairs because $p-2$ has factors $(n-1)$ and $(n+1)$.  Further, $n^2+1 = (n^2-a^2)+(a^2+1)$. If $(n^2+1)$ is prime, then $p-a^2-1$ cannot be for all $0&lt;a&lt;n-1$. For a even, $a^2+1$ is odd and $p-a^2-1$ is even, so only $a$ odd is of interest. Conclude: Given two primes $p_1$ and $p_2$, where $p_2&gt;p_1$, then if $p_2=n^2+1$ for certain $n&gt;0$, then $p_2-p_1$ cannot equal $a^2+1$, $0&lt;a&lt;n-1$.

So the primes $n^2+1$ can produce are somewhat reduced in number because of the growing number of gap conditions that need to be satisfied.  And the further one goes out, the more conditions there are to satisfy. So another form of the $n^2+1$ conjecture might be to ask whether it is possible to satisfy the conditions as $n$ increases.  -FW