I want to prove that $1+i$ is a prime element of ring of Gaussian integers $Z[i]$. I started off with saying that if $1+i$ is prime then $1+i|(a+bi)(c+di)$ implies $1+i|a+bi$ or $1+i|c+di$ for $a+bi,c+di$ in $Z[i]$. Taking conjugate, we get
$1-i|(a-bi)(c-di)$
Next we get,
$2|(a^2+b^2)(c^2+d^2)$
Since $2$ is a prime in $Z$ hence $2|(a^2+b^2)$ or $2|(c^2+d^2)$. After this I am stuck. I do not know how to get $1+i|a+bi$ or $1+i|c+di$ from here. Please suggest.
Firstly, your way of phrasing your solution is not entirely correct. If you are trying to show that $z$ is prime, you should not start with "If $z$ is prime". However, your working is actually salvageable.
You showed that if $(1 + i) \mid (a + bi)(c + di)$, then that alone implies $2 \mid (a^2 + b^2)(c^2 + d^2)$.
(Without any assumption of $1 + i$ being prime. This is what I meant with "salvageable".)
WLOG, we have $2 \mid (a^2 + b^2)$. (We wish to show that $(1 + i) \mid (a + bi)$.)
Thus, either $a$ and $b$ are both even or both odd.
Case 1. $a$ and $b$ are even.
In this case, $2 \mid a$ and $2 \mid b$ and hence, $2 \mid (a + bi)$. Since $1 + i \mid 2$, you are done.
Case 2. $a$ and $b$ are odd.
Write $a = 2a' + 1$ and $b = 2b' + 1$ for integers $a', b'$.
Then, $$a + bi = 2(a' + b'i) + (1 + i).$$ Again, since $(1 + i) \mid 2$, it follows that $(1 + i) \mid a + bi$.