Suppose there are only finitely many primes of the form $25m+7$. Let $p_1,\ldots, p_r$ be such primes and let $M = (5p_1\ldots p_r)^2 + 7$. Then $N$ has form $25m + 7$, and is not divisible by any $p_i$. Let $p$ be any prime dividing $M$. Then $(5p_1\ldots p_r)^2 \equiv -7 \mod{p}$, but this means that the number $−7$ is a square$\mod p$,which is only possible if $p \equiv 7 \mod 25$. This contradicts the fact that no $p_i$ divides $M$. Therefore there must be infinitely of that form.
Does this prove my claim? I was following a layout of a proof for primes of the from $4n+1$, and tried to modify it.
Update: the claim that $-7$ is a square $\mod p$,whenever $p\equiv 7 \mod 25$ does not hold so some other claim is needed
The logical structure of your proof is fine, but there is one huge claim that you have stated without proof:
This claim is in fact not true. For instance, $-7$ is a square mod $2$, or mod $11$, even though $2$ and $11$ are not $7$ mod $25$. More generally, by quadratic reciprocity, if $p\neq 2,7$ is a prime, then $-7$ is a square mod $p$ iff $p$ is a square mod $7$; that is, iff $p$ is $1,2,$ or $4$ mod $7$.